IB Math Analysis & Approaches Questionbank-Topic: SL 3.5 Definition of trigonometric ratios and Exact values SL Paper 1

Question

Let \(p = \sin 40^\circ \) and \(q = \cos 110^\circ \) . Give your answers to the following in terms of p and/or q .

Write down an expression for

(i)     \(\sin 140^\circ \) ;

(ii)    \(\cos 70^\circ \) .

[2]
a(i) and (ii).

Find an expression for \(\cos 140^\circ \) .

[3]
b.

Find an expression for \(\tan 140^\circ \) .

[1]
c.
Answer/Explanation

Markscheme

(i) \(\sin 140^\circ = p\)     A1     N1

(ii) \(\cos 70^\circ = – q\)     A1     N1

[2 marks]

a(i) and (ii).

METHOD 1

evidence of using \({\sin ^2}\theta + {\cos ^2}\theta = 1\)     (M1)

e.g. diagram, \(\sqrt {1 – {p^2}} \) (seen anywhere)

\(\cos 140^\circ = \pm \sqrt {1 – {p^2}} \)     (A1)

\(\cos 140^\circ = – \sqrt {1 – {p^2}} \)     A1     N2

METHOD 2

evidence of using \(\cos 2\theta = 2{\cos ^2}\theta – 1\)     (M1)

\(\cos 140^\circ = 2{\cos ^2}70 – 1\)     (A1)

\(\cos 140^\circ = 2{( – q)^2} – 1\) \(( = 2{q^2} – 1)\)     A1     N2

[3 marks]

b.

METHOD 1

\(\tan 140^\circ = \frac{{\sin 140^\circ }}{{\cos 140^\circ }} = – \frac{p}{{\sqrt {1 – {p^2}} }}\)     A1     N1

METHOD 2

\(\tan 140^\circ = \frac{p}{{2{q^2} – 1}}\)     A1     N1

[1 mark]

c.

Question

A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.


The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is \(\theta \) radians, where \(0 \le \theta  \le \frac{\pi }{2}\) .

Write down an expression in terms of \(\theta \) for

(i)     \(x\) ;

(ii)    \(y\) .

[2]
a.

Let the area of the rectangle be A.

Show that \(A = 18\sin 2\theta \) .

[3]
b.

(i)     Find \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }}\) .

(ii)    Hence, find the exact value of \(\theta \) which maximizes the area of the rectangle.

(iii)   Use the second derivative to justify that this value of \(\theta \) does give a maximum.

[8]
c.
Answer/Explanation

Markscheme

(i) \(x = 3\cos \theta \)     A1     N1 

(ii) \(y = 3\sin \theta \)     A1     N1

[2 marks]

a.

finding area     (M1)

e.g. \(A = 2x \times 2y\) , \(A = 8 \times \frac{1}{2}bh\) 

substituting     A1

e.g. \(A = 4 \times 3\sin \theta  \times 3\cos \theta \) , \(8 \times \frac{1}{2} \times 3\cos \theta  \times 3\sin \theta \)

\(A = 18(2\sin \theta \cos \theta )\)    A1

\(A = 18\sin 2\theta \)     AG     N0

[3 marks]

b.

(i) \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta \)     A2     N2 

(ii) for setting derivative equal to 0     (M1)

e.g. \(36\cos 2\theta  = 0\) , \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 0\)

\(2\theta  = \frac{\pi }{2}\)     (A1)

\(\theta  = \frac{\pi }{4}\)     A1     N2

(iii) valid reason (seen anywhere)     R1

e.g. at \(\frac{\pi }{4}\), \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} < 0\) ; maximum when \(f”(x) < 0\)

finding second derivative \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} =  – 72\sin 2\theta \)     A1

evidence of substituting \(\frac{\pi }{4}\)     M1

e.g. \( – 72\sin \left( {2 \times \frac{\pi }{4}} \right)\) , \( – 72\sin \left( {\frac{\pi }{2}} \right)\) , \( – 72\)

\(\theta  = \frac{\pi }{4}\) produces the maximum area     AG     N0

[8 marks]

c.

Question

Six equilateral triangles, each with side length 3 cm, are arranged to form a hexagon.
This is shown in the following diagram.

The vectors p , q and r are shown on the diagram.

Find p•(p + q + r).

Answer/Explanation

Markscheme

METHOD 1 (using |p| |2q| cosθ)

finding p + q + r (A1)

eg 2q,

| p + q + r | = 2 × 3 (= 6) (seen anywhere) A1

correct angle between p and q (seen anywhere) (A1)

\(\frac{\pi }{3}\) (accept 60°)

substitution of their values (M1)

eg 3 × 6 × cos\(\left( {\frac{\pi }{3}} \right)\)

correct value for cos\(\left( {\frac{\pi }{3}} \right)\) (seen anywhere) (A1)

eg \(\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}\)

p•(p + q + r) = 9 A1 N3

METHOD 2 (scalar product using distributive law)

correct expression for scalar distribution (A1)

eg pp + pq + pr

three correct angles between the vector pairs (seen anywhere) (A2)

eg 0° between p and p, \(\frac{\pi }{3}\) between p and q, \(\frac{{2\pi }}{3}\) between p and r

Note: Award A1 for only two correct angles.

substitution of their values (M1)

eg 3.3.cos0 +3.3.cos\(\frac{\pi }{3}\) + 3.3.cos120

one correct value for cos0, cos\(\left( {\frac{\pi }{3}} \right)\) or cos\(\left( {\frac{2\pi }{3}} \right)\) (seen anywhere) A1

eg \(\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}\)

p•(p + q + r) = 9 A1 N3

METHOD 3 (scalar product using relative position vectors)

valid attempt to find one component of p or r (M1)

eg sin 60 = \(\frac{x}{3}\), cos 60 = \(\frac{x}{3}\), one correct value \(\frac{3}{2},\,\,\frac{{3\sqrt 3 }}{2},\,\,\frac{{ – 3\sqrt 3 }}{2}\)

one correct vector (two or three dimensions) (seen anywhere) A1

eg \(p = \left( \begin{gathered}
\,\,\,\frac{3}{2} \hfill \\
\frac{{3\sqrt 3 }}{2} \hfill \\
\end{gathered} \right),\,\,q = \left( \begin{gathered}
3 \hfill \\
0 \hfill \\
\end{gathered} \right),\,\,r = \left( \begin{gathered}
\,\,\,\,\frac{3}{2} \hfill \\
– \frac{{3\sqrt 3 }}{2} \hfill \\
\,\,\,\,0 \hfill \\
\end{gathered} \right)\)

three correct vectors p + q + r = 2q (A1)

p + q + r = \(\left( \begin{gathered}
6 \hfill \\
0 \hfill \\
\end{gathered} \right)\) or \(\left( \begin{gathered}
6 \hfill \\
0 \hfill \\
0 \hfill \\
\end{gathered} \right)\) (seen anywhere, including scalar product) (A1)

correct working (A1)
eg \(\left( {\frac{3}{2} \times 6} \right) + \left( {\frac{{3\sqrt 3 }}{2} \times 0} \right),\,\,9 + 0 + 0\)

p•(p + q + r) = 9 A1 N3

[6 marks]

Question

The first two terms of an infinite geometric sequence are u1 = 18 and u2 = 12sin2 θ , where 0 < θ < 2\(\pi \) , and θ ≠ \(\pi \).

Find an expression for r in terms of θ.

[2]
a.i.

Find the possible values of r.

[3]
a.ii.

Show that the sum of the infinite sequence is \(\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}\).

[4]
b.

Find the values of θ which give the greatest value of the sum.

[6]
c.
Answer/Explanation

Markscheme

valid approach     (M1)

eg   \(\frac{{{u_2}}}{{{u_1}}},\,\,\frac{{{u_1}}}{{{u_2}}}\)

\(r = \frac{{12\,{{\sin }^2}\,\theta }}{{18}}\left( { = \frac{{2\,{{\sin }^2}\,\theta }}{3}} \right)\)      A1 N2

[2 marks]

a.i.

recognizing that sinθ is bounded      (M1)

eg    0 sin2 θ ≤ 1, −1 ≤ sinθ ≤ 1, −1 < sinθ < 1

0 < r ≤ \(\frac{2}{3}\)      A2 N3

Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).
If no working shown, award N1 for correct values with incorrect inequality sign(s).

[3 marks]

a.ii.

correct substitution into formula for infinite sum       A1

eg  \(\frac{{18}}{{1 – \frac{{2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

evidence of choosing an appropriate rule for cos 2θ (seen anywhere)         (M1)

eg   cos 2θ = 1 − 2 sin2 θ

correct substitution of identity/working (seen anywhere)      (A1)

eg   \(\frac{{18}}{{1 – \frac{2}{3}\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 – 2\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 – 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

correct working that clearly leads to the given answer       A1

eg  \(\frac{{18 \times 3}}{{2 + \left( {1 – 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)}},\,\,\frac{{54}}{{3 – \left( {1 – {\text{cos}}\,2\theta } \right)}}\)

\(\frac{{54}}{{2 + {\text{cos}}\left( {2\theta } \right)}}\)    AG N0

[4 marks]

b.

METHOD 1 (using differentiation)

recognizing \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }} = 0\) (seen anywhere)       (M1)

finding any correct expression for \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }}\)       (A1)

eg  \(\frac{{0 – 54 \times \left( { – 2\,{\text{sin}}\,2\,\theta } \right)}}{{{{\left( {2 + {\text{cos}}\,2\,\theta } \right)}^2}}},\,\, – 54{\left( {2 + {\text{cos}}\,2\,\theta } \right)^{ – 2}}\,\left( { – 2\,{\text{sin}}\,2\,\theta } \right)\)

correct working       (A1)

eg  sin 2θ = 0

any correct value for sin−1(0) (seen anywhere)       (A1)

eg  0, \(\pi \), … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore additional values)      (A1)

2θ = \(\pi \), 3\(\pi \) (accept values in degrees)

both correct answers \(\theta  = \frac{\pi }{2},\,\frac{{3\pi }}{2}\)      A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

METHOD 2 (using denominator)

recognizing when S is greatest      (M1)

eg 2 + cos 2θ is a minimum, 1−r is smallest
correct working      (A1)

eg  minimum value of 2 + cos 2θ is 1, minimum r = \(\frac{2}{3}\)

correct working      (A1)

eg  \({\text{cos}}\,2\,\theta  =  – 1,\,\,\frac{2}{3}\,{\text{si}}{{\text{n}}^2}\,\theta  = \frac{2}{3},\,\,{\text{si}}{{\text{n}}^2}\theta  = 1\)

EITHER (using cos 2θ)

any correct value for cos−1(−1) (seen anywhere)      (A1)

eg  \(\pi \), 3\(\pi \), … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked

both correct values for 2θ  (ignore additional values)      (A1)

2θ = \(\pi \), 3\(\pi \) (accept values in degrees)

OR (using sinθ)

sinθ = ±1     (A1)

sin−1(1) = \(\frac{\pi }{2}\) (accept values in degrees) (seen anywhere)      A1

THEN

both correct answers \(\theta  = \frac{\pi }{2},\,\frac{{3\pi }}{2}\)       A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

[6 marks]

c.

Question

Let \(p = \sin 40^\circ \) and \(q = \cos 110^\circ \) . Give your answers to the following in terms of p and/or q .

Write down an expression for

(i)     \(\sin 140^\circ \) ;

(ii)    \(\cos 70^\circ \) .

[2]
a(i) and (ii).

Find an expression for \(\cos 140^\circ \) .

[3]
b.

Find an expression for \(\tan 140^\circ \) .

[1]
c.
Answer/Explanation

Markscheme

(i) \(\sin 140^\circ = p\)     A1     N1

(ii) \(\cos 70^\circ = – q\)     A1     N1

[2 marks]

a(i) and (ii).

METHOD 1

evidence of using \({\sin ^2}\theta + {\cos ^2}\theta = 1\)     (M1)

e.g. diagram, \(\sqrt {1 – {p^2}} \) (seen anywhere)

\(\cos 140^\circ = \pm \sqrt {1 – {p^2}} \)     (A1)

\(\cos 140^\circ = – \sqrt {1 – {p^2}} \)     A1     N2

METHOD 2

evidence of using \(\cos 2\theta = 2{\cos ^2}\theta – 1\)     (M1)

\(\cos 140^\circ = 2{\cos ^2}70 – 1\)     (A1)

\(\cos 140^\circ = 2{( – q)^2} – 1\) \(( = 2{q^2} – 1)\)     A1     N2

[3 marks]

b.

METHOD 1

\(\tan 140^\circ = \frac{{\sin 140^\circ }}{{\cos 140^\circ }} = – \frac{p}{{\sqrt {1 – {p^2}} }}\)     A1     N1

METHOD 2

\(\tan 140^\circ = \frac{p}{{2{q^2} – 1}}\)     A1     N1

[1 mark]

c.

Question

Let \(f(x) = \sqrt 3 {{\rm{e}}^{2x}}\sin x + {{\rm{e}}^{2x}}\cos x\) , for \(0 \le x \le \pi \) . Solve the equation \(f(x) = 0\) .

Answer/Explanation

Markscheme

\({{\rm{e}}^{2x}}\left( {\sqrt 3 \sin x + \cos x} \right) = 0\)     (A1)

\({{\rm{e}}^{2x}} = 0\) not possible (seen anywhere)     (A1)

simplifying

e.g. \(\sqrt 3 \sin x + \cos x = 0\) , \(\sqrt 3 \sin x =  – \cos x\) , \(\frac{{\sin x}}{{ – \cos x}} = \frac{1}{{\sqrt 3 }}\)     A1 

EITHER

\(\tan x = – \frac{1}{{\sqrt 3 }}\)     A1

\(x = \frac{{5\pi }}{6}\)     A2     N4

OR

sketch of \(30^\circ \) , \(60^\circ \) , \(90^\circ \) triangle with sides \(1\), \(2\), \(\sqrt 3 \)     A1

work leading to \(x = \frac{{5\pi }}{6}\)     A1

verifying \(\frac{{5\pi }}{6}\) satisfies equation     A1     N4

[6 marks]

Question

The first two terms of an infinite geometric sequence are u1 = 18 and u2 = 12sin2 θ , where 0 < θ < 2\(\pi \) , and θ ≠ \(\pi \).

Find an expression for r in terms of θ.

[2]
a.i.

Find the possible values of r.

[3]
a.ii.

Show that the sum of the infinite sequence is \(\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}\).

[4]
b.

Find the values of θ which give the greatest value of the sum.

[6]
c.
Answer/Explanation

Markscheme

valid approach     (M1)

eg   \(\frac{{{u_2}}}{{{u_1}}},\,\,\frac{{{u_1}}}{{{u_2}}}\)

\(r = \frac{{12\,{{\sin }^2}\,\theta }}{{18}}\left( { = \frac{{2\,{{\sin }^2}\,\theta }}{3}} \right)\)      A1 N2

[2 marks]

a.i.

recognizing that sinθ is bounded      (M1)

eg    0 sin2 θ ≤ 1, −1 ≤ sinθ ≤ 1, −1 < sinθ < 1

0 < r ≤ \(\frac{2}{3}\)      A2 N3

Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).
If no working shown, award N1 for correct values with incorrect inequality sign(s).

[3 marks]

a.ii.

correct substitution into formula for infinite sum       A1

eg  \(\frac{{18}}{{1 – \frac{{2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

evidence of choosing an appropriate rule for cos 2θ (seen anywhere)         (M1)

eg   cos 2θ = 1 − 2 sin2 θ

correct substitution of identity/working (seen anywhere)      (A1)

eg   \(\frac{{18}}{{1 – \frac{2}{3}\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 – 2\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 – 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

correct working that clearly leads to the given answer       A1

eg  \(\frac{{18 \times 3}}{{2 + \left( {1 – 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)}},\,\,\frac{{54}}{{3 – \left( {1 – {\text{cos}}\,2\theta } \right)}}\)

\(\frac{{54}}{{2 + {\text{cos}}\left( {2\theta } \right)}}\)    AG N0

[4 marks]

b.

METHOD 1 (using differentiation)

recognizing \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }} = 0\) (seen anywhere)       (M1)

finding any correct expression for \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }}\)       (A1)

eg  \(\frac{{0 – 54 \times \left( { – 2\,{\text{sin}}\,2\,\theta } \right)}}{{{{\left( {2 + {\text{cos}}\,2\,\theta } \right)}^2}}},\,\, – 54{\left( {2 + {\text{cos}}\,2\,\theta } \right)^{ – 2}}\,\left( { – 2\,{\text{sin}}\,2\,\theta } \right)\)

correct working       (A1)

eg  sin 2θ = 0

any correct value for sin−1(0) (seen anywhere)       (A1)

eg  0, \(\pi \), … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore additional values)      (A1)

2θ = \(\pi \), 3\(\pi \) (accept values in degrees)

both correct answers \(\theta  = \frac{\pi }{2},\,\frac{{3\pi }}{2}\)      A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

METHOD 2 (using denominator)

recognizing when S is greatest      (M1)

eg 2 + cos 2θ is a minimum, 1−r is smallest
correct working      (A1)

eg  minimum value of 2 + cos 2θ is 1, minimum r = \(\frac{2}{3}\)

correct working      (A1)

eg  \({\text{cos}}\,2\,\theta  =  – 1,\,\,\frac{2}{3}\,{\text{si}}{{\text{n}}^2}\,\theta  = \frac{2}{3},\,\,{\text{si}}{{\text{n}}^2}\theta  = 1\)

EITHER (using cos 2θ)

any correct value for cos−1(−1) (seen anywhere)      (A1)

eg  \(\pi \), 3\(\pi \), … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked

both correct values for 2θ  (ignore additional values)      (A1)

2θ = \(\pi \), 3\(\pi \) (accept values in degrees)

OR (using sinθ)

sinθ = ±1     (A1)

sin−1(1) = \(\frac{\pi }{2}\) (accept values in degrees) (seen anywhere)      A1

THEN

both correct answers \(\theta  = \frac{\pi }{2},\,\frac{{3\pi }}{2}\)       A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

[6 marks]

c.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

[N/A]

c.

Question

Let \(f(x) = {{\rm{e}}^{ – 3x}}\) and \(g(x) = \sin \left( {x – \frac{\pi }{3}} \right)\) .

Write down

(i)     \(f'(x)\) ;

(ii)    \(g'(x)\) .

[2]
a.

Let \(h(x) = {{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right)\) . Find the exact value of \(h’\left( {\frac{\pi }{3}} \right)\) .

[4]
b.
Answer/Explanation

Markscheme

(i) \( – 3{{\rm{e}}^{ – 3x}}\)     A1     N1

(ii) \(\cos \left( {x – \frac{\pi }{3}} \right)\)     A1     N1

[4 marks]

a.

evidence of choosing product rule     (M1)

e.g. \(uv’ + vu’\)

correct expression     A1

e.g. \( – 3{{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3x}}\cos \left( {x – \frac{\pi }{3}} \right)\)

complete correct substitution of \(x = \frac{\pi }{3}\)     (A1)

e.g. \( – 3{{\rm{e}}^{ – 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right)\)        

\(h’\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ – \pi }}\)     A1     N3

[4 marks]

b.

Question

Let \(f(x) = \frac{{\cos x}}{{\sin x}}\) , for \(\sin x \ne 0\) .

In the following table, \(f’\left( {\frac{\pi }{2}} \right) = p\) and \(f”\left( {\frac{\pi }{2}} \right) = q\) . The table also gives approximate values of \(f'(x)\) and \(f”(x)\) near \(x = \frac{\pi }{2}\) .


Use the quotient rule to show that \(f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}\) .

[5]
a.

Find \(f”(x)\) .

[3]
b.

Find the value of p and of q.

[3]
c.

Use information from the table to explain why there is a point of inflexion on the graph of f where \(x = \frac{\pi }{2}\) .

[2]
d.
Answer/Explanation

Markscheme

\(\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = – \sin x\) (seen anywhere)     (A1)(A1)

evidence of using the quotient rule     M1

correct substitution     A1

e.g. \(\frac{{\sin x( – \sin x) – \cos x(\cos x)}}{{{{\sin }^2}x}}\) , \(\frac{{ – {{\sin }^2}x – {{\cos }^2}x}}{{{{\sin }^2}x}}\)

\(f'(x) = \frac{{ – ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}\)     A1

\(f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}\)     AG      N0

[5 marks]

a.

METHOD 1

appropriate approach     (M1)

e.g. \(f'(x) = – {(\sin x)^{ – 2}}\)

\(f”(x) = 2({\sin ^{ – 3}}x)(\cos x)\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\)     A1A1     N3

Note: Award A1 for \(2{\sin ^{ – 3}}x\) , A1 for \(\cos x\) .

METHOD 2

derivative of \({\sin ^2}x = 2\sin x\cos x\) (seen anywhere)     A1

evidence of choosing quotient rule     (M1)

e.g. \(u = – 1\) ,  \(v = {\sin ^2}x\) , \(f” = \frac{{{{\sin }^2}x \times 0 – ( – 1)2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\)

\(f”(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\)     A1     N3

[3 marks]

b.

evidence of substituting \(\frac{\pi }{2}\)     M1

e.g. \(\frac{{ – 1}}{{{{\sin }^2}\frac{\pi }{2}}}\) , \(\frac{{2\cos \frac{\pi }{2}}}{{{{\sin }^3}\frac{\pi }{2}}}\)

\(p = – 1\) ,  \(q = 0\)    A1A1     N1N1

[3 marks]

c.

second derivative is zero, second derivative changes sign     R1R1     N2

[2 marks]

d.

Question

Let \(f(x) = \cos 2x\) and \(g(x) = 2{x^2} – 1\) .

Find \(f\left( {\frac{\pi }{2}} \right)\) .

[2]
a.

Find \((g \circ f)\left( {\frac{\pi }{2}} \right)\) .

[2]
b.

Given that \((g \circ f)(x)\) can be written as \(\cos (kx)\) , find the value of k, \(k \in \mathbb{Z}\) .

[3]
c.
Answer/Explanation

Markscheme

\(f\left( {\frac{\pi }{2}} \right) = \cos \pi \)     (A1)

\( = – 1\)     A1     N2

[2 marks]

a.

\((g \circ f)\left( {\frac{\pi }{2}} \right) = g( – 1)\) \(( = 2{( – 1)^2} – 1)\)    (A1)

\(= 1\)     A1     N2

[2 marks]

b.

\((g \circ f)(x) = 2{(\cos (2x))^2} – 1\) \(( = 2{\cos ^2}(2x) – 1)\)     A1

evidence of \(2{\cos ^2}\theta – 1 = \cos 2\theta \) (seen anywhere)     (M1)

\((g \circ f)(x) = \cos 4x\)

\(k = 4\)     A1     N2

[3 marks]

c.

Question

Let \(h(x) = \frac{{6x}}{{\cos x}}\) . Find \(h'(0)\) .

Answer/Explanation

Markscheme

METHOD 1 (quotient)

derivative of numerator is 6     (A1)

derivative of denominator is \( – \sin x\)     (A1)

attempt to substitute into quotient rule     (M1)

correct substitution     A1

e.g. \(\frac{{(\cos x)(6) – (6x)( – \sin x)}}{{{{(\cos x)}^2}}}\)

substituting \(x = 0\)     (A1)

e.g. \(\frac{{(\cos 0)(6) – (6 \times 0)( – \sin 0)}}{{{{(\cos 0)}^2}}}\)

\(h'(0) = 6\)     A1     N2

METHOD 2 (product)

\(h(x) = 6x \times {(\cos x)^{ – 1}}\)

derivative of 6x is 6     (A1)

derivative of \({(\cos x)^{ – 1}}\) is \(( – {(\cos x)^{ – 2}}( – \sin x))\)     (A1)

attempt to substitute into product rule     (M1)

correct substitution     A1

e.g. \((6x)( – {(\cos x)^{ – 2}}( – \sin x)) + (6){(\cos x)^{ – 1}}\)

substituting \(x = 0\)    (A1)

e.g. \((6 \times 0)( – {(\cos 0)^{ – 2}}( – \sin 0)) + (6){(\cos 0)^{ – 1}}\)

\(h'(0) = 6\)     A1     N2

[6 marks]

Question

Let \(\int_\pi ^a {\cos 2x{\text{d}}x}  = \frac{1}{2}{\text{, where }}\pi  < a < 2\pi \). Find the value of \(a\).

Answer/Explanation

Markscheme

correct integration (ignore absence of limits and “\(+C\)”)     (A1)

eg     \(\frac{{\sin (2x)}}{2},{\text{ }}\int_\pi ^a {\cos 2x = \left[ {\frac{1}{2}\sin (2x)} \right]_\pi ^a} \)

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg     \(\frac{1}{2}\sin (2a) – \frac{1}{2}\sin (2\pi ),{\text{ }}\sin (2\pi ) – \sin (2a)\)

\(\sin (2\pi ) = 0\)     (A1)

setting their result from an integrated function equal to \(\frac{1}{2}\)     M1

eg     \(\frac{1}{2}\sin 2a = \frac{1}{2},{\text{ }}\sin (2a) = 1\)

recognizing \({\sin ^{ – 1}}1 = \frac{\pi }{2}\)     (A1)

eg     \(2a = \frac{\pi }{2},{\text{ }}a = \frac{\pi }{4}\)

correct value     (A1)

eg     \(\frac{\pi }{2} + 2\pi ,{\text{ }}2a = \frac{{5\pi }}{2},{\text{ }}a = \frac{\pi }{4} + \pi \)

\(a = \frac{{5\pi }}{4}\)     A1     N3

[7 marks]

Question

The following diagram shows a triangle ABC and a sector BDC of a circle with centre B and radius 6 cm. The points A , B and D are on the same line.

M16/5/MATME/SP1/ENG/TZ2/05

\({\text{AB}} = 2\sqrt 3 {\text{ cm, BC}} = 6{\text{ cm, area of triangle ABC}} = 3\sqrt 3{\text{ c}}{{\text{m}}^{\text{2}}}{\rm{, A\hat BC}}\) is obtuse.

Find \({\rm{A\hat BC}}\).

[5]
a.

Find the exact area of the sector BDC.

[3]
b.
Answer/Explanation

Markscheme

METHOD 1

correct substitution into formula for area of triangle     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B,{\text{ }}6\sqrt 3 \sin B,{\text{ }}\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B = 3\sqrt 3 \)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6\sqrt 3 \sin B = 3\sqrt 3 ,{\text{ }}\sin B = \frac{{3\sqrt 3 }}{{\frac{1}{2}(6)2\sqrt 3 }}\)

\(\sin B = \frac{1}{2}\)    (A1)

\(\frac{\pi }{6}(30^\circ )\)    (A1)

\({\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )\)     A1     N3

METHOD 2

(using height of triangle ABC by drawing perpendicular segment from C to AD)

correct substitution into formula for area of triangle     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}\left( {2\sqrt 3 } \right)(h) = 3\sqrt 3 ,{\text{ }}h\sqrt 3 \)

correct working     (A1)

eg\(\,\,\,\,\,\)\(h\sqrt 3  = 3\sqrt 3 \)

height of triangle is 3     A1

\({\rm{C\hat BD}} = \frac{\pi }{6}(30^\circ )\)    (A1)

\({\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )\)     A1     N3

[5 marks]

a.

recognizing supplementary angle     (M1)

eg\(\,\,\,\,\,\)\({\rm{C\hat BD}} = \frac{\pi }{6},{\text{ sector}} = \frac{1}{2}(180 – {\rm{A\hat BC)(}}{{\text{6}}^2})\)

correct substitution into formula for area of sector     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2} \times \frac{\pi }{6} \times {6^2},{\text{ }}\pi ({6^2})\left( {\frac{{30}}{{360}}} \right)\)

\({\text{area}} = 3\pi {\text{ }}({\text{c}}{{\text{m}}^2})\)     A1     N2

[3 marks]

b.

Question

The following diagram shows triangle PQR.

M17/5/MATME/SP1/ENG/TZ1/03

Find PR.

Answer/Explanation

Markscheme

METHOD 1 

evidence of choosing the sine rule     (M1)

eg\(\,\,\,\,\,\)\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}}\)

correct substitution     A1

eg\(\,\,\,\,\,\)\(\frac{x}{{\sin 30}} = \frac{{13}}{{\sin 45}},{\text{ }}\frac{{13\sin 30}}{{\sin 45}}\)

\(\sin 30 = \frac{1}{2},{\text{ }}\sin 45 = \frac{1}{{\sqrt 2 }}\)     (A1)(A1)

correct working     A1

eg\(\,\,\,\,\,\)\(\frac{1}{2} \times \frac{{13}}{{\frac{1}{{\sqrt 2 }}}},{\text{ }}\frac{1}{2} \times 13 \times \frac{2}{{\sqrt 2 }},{\text{ }}13 \times \frac{1}{2} \times \sqrt 2 \)

correct answer     A1     N3

eg\(\,\,\,\,\,\)\({\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}\)

METHOD 2 (using height of ΔPQR)

valid approach to find height of ΔPQR     (M1)

eg\(\,\,\,\,\,\)\(\sin 30 = \frac{x}{{13}},{\text{ }}\cos 60 = \frac{x}{{13}}\)

\(\sin 30 = \frac{1}{2}{\text{ or }}\cos 60 = \frac{1}{2}\)     (A1)

\({\text{height}} = 6.5\)     A1

correct working     A1

eg\(\,\,\,\,\,\)\(\sin 45 = \frac{{6.5}}{{{\text{PR}}}},{\text{ }}\sqrt {{{6.5}^2} + {{6.5}^2}} \)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\sin 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\cos 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\sqrt {\frac{{169 \times 2}}{4}} \)

correct answer     A1     N3

eg\(\,\,\,\,\,\)\({\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}\)

[6 marks]

Question

The following table shows the probability distribution of a discrete random variable \(A\), in terms of an angle \(\theta \).

M17/5/MATME/SP1/ENG/TZ1/10

Show that \(\cos \theta  = \frac{3}{4}\).

[6]
a.

Given that \(\tan \theta  > 0\), find \(\tan \theta \).

[3]
b.

Let \(y = \frac{1}{{\cos x}}\), for \(0 < x < \frac{\pi }{2}\). The graph of \(y\)between \(x = \theta \) and \(x = \frac{\pi }{4}\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.

[6]
c.
Answer/Explanation

Markscheme

evidence of summing to 1     (M1)

eg\(\,\,\,\,\,\)\(\sum {p = 1} \)

correct equation     A1

eg\(\,\,\,\,\,\)\(\cos \theta  + 2\cos 2\theta  = 1\)

correct equation in \(\cos \theta \)     A1

eg\(\,\,\,\,\,\)\(\cos \theta  + 2(2{\cos ^2}\theta  – 1) = 1,{\text{ }}4{\cos ^2}\theta  + \cos \theta  – 3 = 0\)

evidence of valid approach to solve quadratic     (M1)

eg\(\,\,\,\,\,\)factorizing equation set equal to \(0,{\text{ }}\frac{{ – 1 \pm \sqrt {1 – 4 \times 4 \times ( – 3)} }}{8}\)

correct working, clearly leading to required answer     A1

eg\(\,\,\,\,\,\)\((4\cos \theta  – 3)(\cos \theta  + 1),{\text{ }}\frac{{ – 1 \pm 7}}{8}\)

correct reason for rejecting \(\cos \theta  \ne  – 1\)     R1

eg\(\,\,\,\,\,\)\(\cos \theta \) is a probability (value must lie between 0 and 1), \(\cos \theta  > 0\)

Note:     Award R0 for \(\cos \theta  \ne  – 1\) without a reason.

\(\cos \theta  = \frac{3}{4}\)    AG  N0

a.

valid approach     (M1)

eg\(\,\,\,\,\,\)sketch of right triangle with sides 3 and 4, \({\sin ^2}x + {\cos ^2}x = 1\)

correct working     

(A1)

eg\(\,\,\,\,\,\)missing side \( = \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}\)

\(\tan \theta  = \frac{{\sqrt 7 }}{3}\)     A1     N2

[3 marks]

b.

attempt to substitute either limits or the function into formula involving \({f^2}\)     (M1)

eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} } \)

correct substitution of both limits and function     (A1)

eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x} \)

correct integration     (A1)

eg\(\,\,\,\,\,\)\(\tan x\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(\tan \frac{\pi }{4} – \tan \theta \)

Note:     Award M0 if they substitute into original or differentiated function.

\(\tan \frac{\pi }{4} = 1\)    (A1)

eg\(\,\,\,\,\,\)\(1 – \tan \theta \)

\(V = \pi  – \frac{{\pi \sqrt 7 }}{3}\)     A1     N3

[6 marks]

c.

Question

The first two terms of an infinite geometric sequence are u1 = 18 and u2 = 12sin2 θ , where 0 < θ < 2\(\pi \) , and θ ≠ \(\pi \).

Find an expression for r in terms of θ.

[2]
a.i.

Find the possible values of r.

[3]
a.ii.

Show that the sum of the infinite sequence is \(\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}\).

[4]
b.

Find the values of θ which give the greatest value of the sum.

[6]
c.
Answer/Explanation

Markscheme

valid approach     (M1)

eg   \(\frac{{{u_2}}}{{{u_1}}},\,\,\frac{{{u_1}}}{{{u_2}}}\)

\(r = \frac{{12\,{{\sin }^2}\,\theta }}{{18}}\left( { = \frac{{2\,{{\sin }^2}\,\theta }}{3}} \right)\)      A1 N2

[2 marks]

a.i.

recognizing that sinθ is bounded      (M1)

eg    0 sin2 θ ≤ 1, −1 ≤ sinθ ≤ 1, −1 < sinθ < 1

0 < r ≤ \(\frac{2}{3}\)      A2 N3

Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).
If no working shown, award N1 for correct values with incorrect inequality sign(s).

[3 marks]

a.ii.

correct substitution into formula for infinite sum       A1

eg  \(\frac{{18}}{{1 – \frac{{2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

evidence of choosing an appropriate rule for cos 2θ (seen anywhere)         (M1)

eg   cos 2θ = 1 − 2 sin2 θ

correct substitution of identity/working (seen anywhere)      (A1)

eg   \(\frac{{18}}{{1 – \frac{2}{3}\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 – 2\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 – 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)

correct working that clearly leads to the given answer       A1

eg  \(\frac{{18 \times 3}}{{2 + \left( {1 – 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)}},\,\,\frac{{54}}{{3 – \left( {1 – {\text{cos}}\,2\theta } \right)}}\)

\(\frac{{54}}{{2 + {\text{cos}}\left( {2\theta } \right)}}\)    AG N0

[4 marks]

b.

METHOD 1 (using differentiation)

recognizing \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }} = 0\) (seen anywhere)       (M1)

finding any correct expression for \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }}\)       (A1)

eg  \(\frac{{0 – 54 \times \left( { – 2\,{\text{sin}}\,2\,\theta } \right)}}{{{{\left( {2 + {\text{cos}}\,2\,\theta } \right)}^2}}},\,\, – 54{\left( {2 + {\text{cos}}\,2\,\theta } \right)^{ – 2}}\,\left( { – 2\,{\text{sin}}\,2\,\theta } \right)\)

correct working       (A1)

eg  sin 2θ = 0

any correct value for sin−1(0) (seen anywhere)       (A1)

eg  0, \(\pi \), … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore additional values)      (A1)

2θ = \(\pi \), 3\(\pi \) (accept values in degrees)

both correct answers \(\theta  = \frac{\pi }{2},\,\frac{{3\pi }}{2}\)      A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

METHOD 2 (using denominator)

recognizing when S is greatest      (M1)

eg 2 + cos 2θ is a minimum, 1−r is smallest
correct working      (A1)

eg  minimum value of 2 + cos 2θ is 1, minimum r = \(\frac{2}{3}\)

correct working      (A1)

eg  \({\text{cos}}\,2\,\theta  =  – 1,\,\,\frac{2}{3}\,{\text{si}}{{\text{n}}^2}\,\theta  = \frac{2}{3},\,\,{\text{si}}{{\text{n}}^2}\theta  = 1\)

EITHER (using cos 2θ)

any correct value for cos−1(−1) (seen anywhere)      (A1)

eg  \(\pi \), 3\(\pi \), … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked

both correct values for 2θ  (ignore additional values)      (A1)

2θ = \(\pi \), 3\(\pi \) (accept values in degrees)

OR (using sinθ)

sinθ = ±1     (A1)

sin−1(1) = \(\frac{\pi }{2}\) (accept values in degrees) (seen anywhere)      A1

THEN

both correct answers \(\theta  = \frac{\pi }{2},\,\frac{{3\pi }}{2}\)       A1 N4

Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.

[6 marks]

c.

Question

The expression \(6\sin x\cos x\) can be expressed in the form \(a\sin bx\) .

Find the value of a and of b .

[3]
a.

Hence or otherwise, solve the equation \(6\sin x\cos x = \frac{3}{2}\) , for \(\frac{\pi }{4} \le x \le \frac{\pi }{2}\) .

[4]
b.
Answer/Explanation

Markscheme

recognizing double angle     M1

e.g. \(3 \times 2\sin x\cos x\) , \(3\sin 2x\)

\(a = 3\) , \(b = 2\)     A1A1     N3

[3 marks]

a.

substitution \(3\sin 2x = \frac{3}{2}\)     M1

\(\sin 2x = \frac{1}{2}\)     A1

finding the angle     A1

e.g. \(\frac{\pi }{6}\) , \(2x = \frac{{5\pi }}{6}\)

\(x = \frac{{5\pi }}{{12}}\)     A1     N2

Note: Award A0 if other values are included.

[4 marks]

b.

Question

Let \(f(x) = \cos x + \sqrt 3 \sin x\) , \(0 \le x \le 2\pi \) . The following diagram shows the graph of \(f\) .


The \(y\)-intercept is at (\(0\), \(1\)) , there is a minimum point at A (\(p\), \(q\)) and a maximum point at B.

Find \(f'(x)\) .

[2]
a.

Hence

(i)     show that \(q = – 2\) ;

(ii)    verify that A is a minimum point.

[10]
b(i) and (ii).

Find the maximum value of \(f(x)\) .

[3]
c.

The function \(f(x)\) can be written in the form \(r\cos (x – a)\) .

Write down the value of r and of a .

[2]
d.
Answer/Explanation

Markscheme

\(f'(x) = – \sin x + \sqrt 3 \cos x\)     A1A1     N2

[2 marks]

a.

(i) at A, \(f'(x) = 0\)     R1

correct working     A1

e.g. \(\sin x = \sqrt 3 \cos x\)

\(\tan x = \sqrt 3 \)     A1

\(x = \frac{\pi }{3}\) , \(\frac{{4\pi }}{3}\)     A1

attempt to substitute their x into \(f(x)\)     M1

e.g. \(\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)\)

correct substitution     A1

e.g. \( – \frac{1}{2} + \sqrt 3 \left( { – \frac{{\sqrt 3 }}{2}} \right)\)

correct working that clearly leads to \( – 2\)     A1

e.g. \( – \frac{1}{2} – \frac{3}{2}\)

 \(q = – 2\)     AG     N0

(ii) correct calculations to find \(f'(x)\) either side of \(x = \frac{{4\pi }}{3}\)     A1A1

e.g. \(f'(\pi ) = 0 – \sqrt 3 \) ,  \(f'(2\pi ) = 0 + \sqrt 3 \)   

\(f'(x)\) changes sign from negative to positive     R1

so A is a minimum     AG     N0

[10 marks]

b(i) and (ii).

max when \(x = \frac{\pi }{3}\)     R1

correctly substituting \(x = \frac{\pi }{3}\) into \(f(x)\)     A1

e.g. \(\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)\)

max value is 2     A1     N1

[3 marks]

c.

\(r = 2\) , \(a = \frac{\pi }{3}\)     A1A1     N2

[2 marks]

d.

MAA SL 3.5-3.6 TRIGONOMETRIC IDENTITIES AND EQUATIONS [concise]-manav

Question

[Maximum mark: 7] [without GDC]
Given that sin x = p , where x is an acute angle
(a) Find the value of cos x in terms of p ; [2]
(b) Hence, express the following in terms of p :
(i) tan x   (ii) cos 2x   (iii) sin 2x   (iv) tan 2x   (v) sin 4x  [5]

Answer/Explanation

Ans:
(a) \(\cos x=\sqrt{1-p^{2}}\)
(b) \(\tan x=\frac{p}{\sqrt{1-p^{2}}}, \quad \cos 2 x=1-2 p^{2}\)
\(\sin 2 x=2 p \sqrt{1-p^{2}}, \quad \tan 2 x=\frac{2 p \sqrt{1-p^{2}}}{1-2 p^{2}}\)
\(\sin 4 x=4 p \sqrt{1-p^{2}}\left(1-2 p^{2}\right)\)

Question

[Maximum mark: 17] [without GDC]
Given that sin 20° = p , and cos 20° = q (so that \(\left.p^{2}+q^{2}=1\right)\) [5]
(a) 
(b) By observing the unit circle   [12]

express the following in terms of p and/or q

Answer/Explanation

Ans:
(a) \(\tan 20^{\circ}=\frac{p}{q}, \quad \sin 40^{\circ}=2 p q, \quad \cos 40^{\circ}=q^{2}-p^{2}=1-2 p^{2}=2 q^{2}-1\)
(b)

Question

[Maximum mark: 4] [without GDC]
Given that \(\sin \theta=\frac{1}{2}, \quad \cos \theta=-\frac{\sqrt{3}}{2} \text { and } 0^{\circ}<\theta<360^{\circ}\)
(a) find the value of θ ; [2]
(b) write down the exact value of tanθ . [2]

Answer/Explanation

Ans:
(a) Acute angle \(30^{\circ} \Rightarrow \theta=150^{\circ}\) (2nd quadrant since sine positive and cosine negative)
(b) \(\tan 150^{\circ}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}\)

Question

[Maximum mark: 6] [without GDC]
Given that \(\sin x=\frac{1}{3}\), where x is an acute angle, find the exact value of
(a) cos x ; [4]
(b) cos 2x . [2]

Answer/Explanation

Ans:
(a) x is an acute angle => cosx is positive.
\(\cos ^{2} x+\sin ^{2} x=1=>\cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=\sqrt{\frac{8}{9}}\left(=\frac{2 \sqrt{2}}{3}\right)\)
(b) \(\cos 2 x=1-2 \sin ^{2} x=1-2\left(\frac{1}{3}\right)^{2}=\frac{7}{9}\)

Question

[Maximum mark: 6] [without GDC]
The following diagram shows a triangle ABC, where ACB is 90°, AB = 3, AC = 2 and BAC is  θ.(a) Show that \(\sin \theta=\frac{\sqrt{5}}{3}\)  [1]
(b) Show that  \(\sin 2 \theta=\frac{4 \sqrt{5}}{9}\) [2]
(c) Find the exact value of cos 2θ. [3]

Answer/Explanation

Ans:
(a) \(\sin \theta=\frac{\mathrm{BC}}{\mathrm{AB}}, \mathrm{BC}=\sqrt{3^{2}-2^{2}}=\sqrt{5} \quad \sin \theta=\frac{\sqrt{5}}{3}\)
(b) \(\sin 2 \theta=2 \sin \theta \cos \theta=2\left(\frac{\sqrt{5}}{3}\right)\left(\frac{2}{3}\right)=\frac{4 \sqrt{5}}{9}\)
(c) \(\cos 2 \theta=\frac{4}{9}-\frac{5}{9}=-\frac{1}{9} \quad \text { OR } \quad \cos 2 \theta=1-2 \times \frac{5}{9}=-\frac{1}{9}\)

Question

[Maximum mark: 4] [without GDC]
If A is an obtuse angle in a triangle and \(\sin A=\frac{5}{13}\), calculate the exact value of sin 2A.

Answer/Explanation

Ans:
\(\sin A=\frac{5}{13} \Rightarrow \cos A=\pm \frac{12}{13}\)  But A is obtuse \(\Rightarrow \cos A=-\frac{12}{13}\)
\(\sin 2 A=2 \sin A \cos A=2 \times \frac{5}{13} \times\left(-\frac{12}{13}\right)=-\frac{120}{169}\)

Question

[Maximum mark: 6] [without GDC]
Let p = sin 40° , q = cos110°. Give your answers to the following in terms of p and/or q
(a) Write down an expression for (i) sin140° ; (ii) cos 70°. [2]
(b) Find an expression for cos140°. [3]
(c) Find an expression for tan140°. [1]

Answer/Explanation

Ans:
(a) (i) sin 140° = p (ii) cos 70° = -q
(b) METHOD 1
using \(\sin ^{2} \theta+\cos ^{2} \theta=1\)
\(\cos 140^{\circ}=\pm \sqrt{1-p^{2}} \quad \cos 140^{\circ}=-\sqrt{1-p^{2}}\)
METHOD 2
\(\text { using } \cos ^{2} \theta=2 \cos ^{2} \theta-1\)
\(\cos 140^{\circ}=2 \cos ^{2} 70-1=2(-q)^{2}-1=2 q^{2}-1\)
(c) METHOD 1  \(\tan 140^{\circ}=\frac{\sin 140^{\circ}}{\cos 140^{\circ}}=-\frac{p}{\sqrt{1-p^{2}}}\)
METHOD 2 \(\tan 140^{\circ}=\frac{p}{2 q^{2}-1}\)

Question

[Maximum mark: 6] [without GDC]
(a) Given that \(\cos A=\frac{1}{3} \text { and } 0 \leq A \leq \frac{\pi}{2}\),  find cos 2A.  [3]
(b) Given that \(\sin B=\frac{2}{3} \text { and } \frac{\pi}{2} \leq B \leq \pi\), find cos B.  [3]

Answer/Explanation

Ans:
(a) \(\cos 2 A=2 \cos ^{2} A-1=2 \times\left(\frac{1}{3}\right)^{2}-1 !=-\frac{7}{9}\)
(b) METHOD 1
using \(\sin ^{2} B+\cos ^{2} B=1\)
\(\cos B=\pm \sqrt{\frac{5}{9}}\left(=\pm \frac{\sqrt{5}}{3}\right)\)    \(\cos B==-\frac{\sqrt{5}}{3}\)
METHOD 2

Diagram, eg
third side equals \(\sqrt{5}\)
\(\cos B=-\frac{\sqrt{5}}{3}\)

Question

[Maximum mark: 7] [without GDC]
The straight line with equation \(y=\frac{3}{4}\)makes an acute angle θ with the x-axis.
(a) Write down the value of tanθ [1]
(b) Find the value of (i) sin 2θ ; (ii) cos 2θ. [6]

Answer/Explanation

Ans:
(a) \(\tan \theta=\frac{3}{4}\)
(b) (i) by using a right-angles triangle with sides 3,4,5
\(\sin \theta=\frac{3}{5}, \cos \theta=\frac{4}{5}\)
\(  \sin 2 \theta=2 \sin x \cos x=\frac{24}{25}\)
(ii)  \(\cos 2 \theta=1-2\left(\frac{3}{5}\right)^{2}=\frac{7}{25} \quad \text { OR } \quad \cos 2 \theta=\left(\frac{3}{5}\right)^{2},\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}=\frac{7}{25}\)

Question

[Maximum mark: 7] [without GDC]
Let \(f(x)=\sin ^{3} x+\cos ^{3} x \tan x, \quad \frac{\pi}{2}<x<\pi\).
(a) Show that \(f(x)=\sin x\) [2]
(b) Let \(\sin x=\frac{2}{3}\). Show that \(f(2 x)=-\frac{4 \sqrt{5}}{9}\). [5]

Answer/Explanation

Ans:
(a) \(f(x)=\sin ^{3} x+\cos ^{3} x \frac{\sin x}{\cos x}=\sin x\left(\sin ^{2} x+\cos ^{2} x\right)=\sin x\)
(b) \(f(2 x)=\sin 2 x=2 \sin x \cos x\)
\(\sin ^{2} x+\cos ^{2} x=1 \Rightarrow \cos x=-\frac{\sqrt{5}}{3}\)
\(   f(2 x)=2\left(\frac{2}{3}\right)\left(-\frac{\sqrt{5}}{3}\right)=-\frac{4 \sqrt{5}}{9} f(2 x)\)

Question

[Maximum mark: 28] [without GDC]
Complete the solutions of the following equations, within the given domain.

Answer/Explanation

Ans:

Question

[Maximum mark: 6] [without GDC]
Solve the equation, \(\sin x=-\frac{1}{2}\), (i) for \(0^{\circ} \leq x \leq 360^{\circ}(\mathrm{deg})\) (ii) \(\text { for } 0 \leq x \leq 2 \pi(\mathrm{rad})\)

Answer/Explanation

Ans:
(i) x = 210°, x = 300°
(ii) x = 7π/6 , x =11π/6

Question

[Maximum mark: 32] [without GDC]
Write down the solutions of the following equations, within the given domain.

Answer/Explanation

Ans:

Question

[Maximum mark: 6] [without GDC]
Solve the equation \(\cos x=\frac{1}{2}\) (i) for \(0^{\circ} \leq x \leq 360^{\circ}(\mathrm{deg})\)  (ii) for \(0 \leq x \leq 2 \pi(\mathrm{rad})\)

Answer/Explanation

Ans:
(i) x = 60° ,x = 300°      (ii) x = π/3 , x = 5π/3

Question

[Maximum mark: 6] [without GDC]
Solve the equation \(\cos x=-\frac{1}{2}\)  (i) for \(0^{\circ} \leq x \leq 360^{\circ}(\mathrm{deg})\)  (ii) for \(0 \leq x \leq 2 \pi(\mathrm{rad})\)

Answer/Explanation

Ans:
(i) x = 120° ,x = 240°      (ii) x = 2π/3 , x = 4π/3

Question

[Maximum mark: 24] [without GDC]
Write down the solutions of the following equations, within the given domain.

Answer/Explanation

Ans:

Question

[Maximum mark: 24] [without GDC]
Write down the solutions of the following equations, within the given domain.

Answer/Explanation


Ans:

Question

[Maximum mark: 7] [without GDC]
Solve the equation \(2 \cos x=\sin 2 x \text {, for } 0 \leq x \leq 3 \pi\).

Answer/Explanation

Ans:
METHOD 1
\(2 \cos x=\sin 2 x \Leftrightarrow 2 \cos x=2 \sin x \cos x \Leftrightarrow 2 \cos x(1-\sin x)=0 \Leftrightarrow \cos x=0 \text { OR } \sin x=1\)
\(x=\frac{\pi}{2}, x=\frac{3 \pi}{2}, x=\frac{5 \pi}{2}\)
intersection points at \(x=\frac{\pi}{2}, x=\frac{3 \pi}{2}, x \frac{5 \pi}{2}\)

Question

[Maximum mark: 6] [with / without GDC]
Solve the equation \(2 \cos ^{2} x=\sin 2 x \text { for } 0 \leq x \leq \pi\), giving your answers in terms of π.

Answer/Explanation

Ans:
\(2 \cos ^{2} x=2 \sin x \cos x \Leftrightarrow 2 \cos ^{2} x-2 \sin x \cos x=0 \Leftrightarrow 2 \cos x(\cos x-\sin x)=0\)
cos x = 0, (cos x – sin x) = 0
cos x = 0 \(\Leftrightarrow x=\frac{\pi}{2}\)
\(\cos x-\sin x=0 \Leftrightarrow \cos x=\sin x \Leftrightarrow \tan x=1 \Leftrightarrow x=\frac{\pi}{4}\)

Question

[Maximum mark: 6] [with GDC]
(a) Factorize the expression \(3 \sin ^{2} x-11 \sin x+6\). [2]
(b) Consider the equation \(3 \sin ^{2} x-11 \sin x+6=0\).
(i) Find the two values of sin x which satisfy this equation,
(ii) Solve the equation, for \(0^{\circ} \leq x \leq 180^{\circ}\). [4]

Answer/Explanation

Ans:
(a) (3 sin x – 2)(sin x – 3)
(b) (i) (3sinx – 2)(sinx – 3) = 0 \(\Rightarrow\) sin x \(\frac{2}{3}\)      sin x = 3
(ii) x = 41.8°, 138°

Question

[Maximum mark: 4] [with GDC]
(a) Write the expression \(3 \sin ^{2} x+4 \cos x\) in the form \(a \cos ^{2} x+b \cos x+c\).  [1]
(b) Hence or otherwise, solve the equation \(3 \sin ^{2} x+4 \cos x-4=0,0^{\circ} \leq x \leq 90^{\circ}\) [3]

Answer/Explanation

Ans:
(a) \(3 \sin ^{2} x+4 \cos x=3\left(1-\cos ^{2} x\right)+4 \cos x=3-3 \cos ^{2}+4 \cos x\)
(b) \(3 \sin ^{2} x+4 \cos x-4=0 \Rightarrow 3-3 \cos ^{2} x+4 \cos x-4=0\)
\(\Rightarrow 3 \cos ^{2} x-4 \cos x+1=0\)
\(\cos x=\frac{1}{3} \text { or } \cos x=1\)
x = 70.5° or x = 0

Question

[Maximum mark: 6] [without GDC]
(a) Express 2cos2 + sin x  in terms of sin x only. [2]
(b) Solve the equation 2cos2x + sinx = 2 for x in the interval \(0 \leq x \leq \pi\)  [4]

Answer/Explanation

Ans:
(a) \(2 \cos ^{2} x+\sin x=2\left(1-\sin ^{2} x\right)+\sin x=2-2 \sin ^{2} x+\sin x\)
(b) \(2 \cos ^{2} x+\sin x=2 \Rightarrow 2-2 \sin ^{2} x+\sin x=2\)
\(\sin x-2 \sin ^{2} x=0\)
sin x(1 – 2sin x) = 0
sin x = 0 or sin x = \(\frac{1}{2}\)
sin x = 0 ⇒ x = 0 or π
\(\sin x=\frac{1}{2} \Rightarrow x=\frac{\pi}{6} \text { or } \frac{5 \pi}{6}\)

Question

[Maximum mark: 6] [without GDC]
Consider the trigonometric equation \(2 \sin ^{2} x=1+\cos x\).
(a) Write this equation in the form f (x) = 0, where \(f(x)=a \sin ^{2} x+b \sin x+c\), and \(a, b, c \in \mathbb{Z}\)   [2]
(b) Factorize f (x) [1]
(c) Solve \(f(x)=0 \text { for } 0^{\circ} \leq x \leq 360^{\circ}\) [3]

Answer/Explanation

Ans:
(a) \(2 \sin ^{2} x=1+\cos x \Leftrightarrow 2\left(1-\cos ^{2} x\right)=1+\cos x \Leftrightarrow 2-2 \cos ^{2} x=1+\cos x\)
\( \Rightarrow 2 \cos ^{2} x+\cos x-1=0\)
(b) \(2 \cos ^{2} x+\cos x-1=(2 \cos x-1)(\cos x+1)\)
(c) \(\cos x=\frac{1}{2} \Rightarrow x=60^{\circ}, \text { or } x=300^{\circ}\)
cos x = –l ⇒ x = 180°
Solutions ⇒ x = 60°, x = 180° , x = 300°

Question

[Maximum mark: 10] [without GDC]
Consider the equation 3cos 2x + sinx = 1.
(a) Write this equation in the form \(f(x)=0, \text { where } f(x)=p \sin ^{2} x+q \sin x+r\) and \(p, q, r \in \mathbb{Z}\).  [2]
(b) Factorize f (x). [2]
(c) Write down the number of solutions of \(f(x)=0, \text { for } 0 \leq x \leq 2 \pi\). [2]
Extra Questions:
(d) Find the two solutions in the 3rd and fourth quadrant [4]

Answer/Explanation

Ans:
(a) \(3 \cos 2 x+\sin x=1 \Leftrightarrow 3\left(1-2 \sin ^{2} x\right)+\sin x=1\)
6 sinx2 – sinx – 2 = 0
(b) (3 sin x – 2)(2 sin x + 1)
(c) 4 solutions
Extra Question
(d) The last equation gives \(\sin x=\frac{2}{3} \text { or } \sin x=-\frac{1}{2}\)
In the 3rd and 4th quadrants sin x is negative. Hence \(\sin x=-\frac{1}{2}\)
\(x=\frac{\pi}{6}+\pi=\frac{7 \pi}{6} \quad \text { or } x=2 \pi-\frac{\pi}{6}=\frac{11 \pi}{6}\)

Question

[Maximum mark: 7] [with / without GDC]
Solve \(\cos 2 x-3 \cos x-3-\cos ^{2} x=\sin ^{2} x, \text { for } 0 \leq x \leq 2 \pi\).

Answer/Explanation

Ans:
\(\cos 2 x-3 \cos x-3-\cos ^{2} x=\sin ^{2} x \Leftrightarrow \cos 2 x-3 \cos x-3=\cos ^{2} x+\sin ^{2} x\)
\(2 \cos ^{2} x-1-3 \cos x-3=1 \Leftrightarrow 2 \cos ^{2} x-3 \cos x-5=0\)
\(\cos x=\frac{5}{2}(\text { rejected }), \cos x=-1\)
\(x=\pi\)

Question

[Maximum mark: 7] [without GDC]
(a) Show that \(4-\cos 2 \theta+5 \sin \theta \equiv 2 \sin ^{2} \theta+5 \sin \theta+3\) [2]
(b) Hence, solve the equation \(4-\cos 2 \theta+5 \sin \theta=0 \text { for } 0 \leq \theta \leq 2 \pi \text {. }\) [5]

Answer/Explanation

Ans:
(a) \(4-\cos 2 \theta+5 \sin \theta=4-\left(1-2 \sin ^{2} \theta\right)+5 \sin \theta=2 \sin ^{2} \theta+5 \sin \theta+3\)
(b) \(\left.2 \sin ^{2} \theta+5 \sin \theta+3=0 \Leftrightarrow 2 \sin \theta+3\right)(\sin \theta+1)=0\)
\(  \sin \theta=-1, ; \sin \theta=-\frac{3}{2} \text { (rejected) }\)
\( \theta=\frac{3 \pi}{2}\)

Question

[Maximum mark: 4] [without GDC]
Solve the equation \(\sqrt{3} \cos x=\sin x\) for x in the interval \(0^{\circ} \leq x \leq 360^{\circ}\)

Answer/Explanation

Ans:
\(\sqrt{3} \cos x=\sin x \Rightarrow \tan x=\sqrt{3} \Rightarrow x=60^{\circ} \text { or } x=240^{\circ}\)

Question

[Maximum mark: 6] [without GDC]
Let \(f(x)=\sqrt{3} \mathrm{e}^{2 x} \sin x+\mathrm{e}^{2 x} \cos x, \text { for } 0 \leq x \leq \pi\). Given that \(\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\), solve the equation f (x) = 0.

Answer/Explanation

Ans:
\(\mathrm{e}^{2 x}(\sqrt{3} \sin x+\cos x)=0\)
\(\mathrm{e}^{2 x}=0 \text { not possible }\)
\(\sqrt{3} \sin x+\cos x=0: \Leftrightarrow \sqrt{3} \sin x=-\cos x: \Leftrightarrow \frac{\sin x}{-\cos x}=\frac{1}{\sqrt{3}} \Leftrightarrow \tan x=-\frac{1}{\sqrt{3}}\)
\(x=\frac{5 \pi}{6}\)

Question

[Maximum mark: 6] [with GDC]
Solve the equations
(a) 3cosx = 5sinx , for x in the interval \(0^{\circ} \leq x \leq 360^{\circ}\), (to the nearest degree) [3]
(b) \(\cos 3 x=\cos (0.5 x), \text { for } 0 \leq x \leq \pi\) [3]
Solve also [without GDC] by using the following information
For (a) it is given that \(\tan ^{-1} 0.6=31^{\circ}\)
For (b) it is given that \(\cos A=\cos B \Rightarrow\left\{\begin{array}{l}
A=B+2 k \pi \\
\text { or } \\
A=-B+2 k \pi
\end{array} \quad \text { for some } k \in Z\right.\)

Answer/Explanation

Ans:
(a) \(x=31^{\circ} \text { or } x=211^{\circ}\)
(b) x= 0, x = 1.80 , x = 2.51 [3 s.f.]
Extra Questions:
(a) \(3 \cos x=5 \sin x \Leftrightarrow \tan x=3 / 5\left(=0.6 \Leftrightarrow x=31^{\circ}+180^{\circ} k\right.\)
\(x=31^{\circ} \text { or } x=211^{\circ}\)
(b) \(\cos 3 x=\cos (0.5 x) \Leftrightarrow 3 x=0.5 x+2 k \pi \text { or } 3 x=-0.5 x+2 k \pi\)
\(3 x=0.5 x+2 k \pi \Leftrightarrow 2.5 x=2 k \pi \Leftrightarrow x=(4 / 5) k \pi \text {. So } x=0, x=(4 / 5) \pi\)
\(3 x=-0.5 x+2 k \pi \Leftrightarrow 3.5 x=2 k \pi \Leftrightarrow x=(4 / 7) k \pi \text {. So } x=0, x=(4 / 7) \pi\)
Solutions: x = 0, x = (4/5)π, x = (4/7)π

Question

[Maximum mark: 4] [without GDC]
Solve the equation \(3 \sin ^{2} x=\cos ^{2} x, \text { for } 0^{\circ} \leq x \leq 180^{\circ} \text {. }\)
[ Please solve in 3 different ways! ]
By expressing the equation in terms of sin x only:
By expressing the equation in terms of cos x only:
By expressing the equation in terms of tan x only:

Answer/Explanation

Ans:
– By expressing the equation in terms of sin x only:
\(3 \sin ^{2} x=\cos ^{2} x \Leftrightarrow 3 \sin ^{2} x=1-\sin ^{2} x \Leftrightarrow 4 \sin ^{2} x=1 \Leftrightarrow \sin ^{2} x=1 / 4 \Leftrightarrow \sin x=\pm 1 / 2\)
Solutions: x = 30° or x = 150°
– By expressing the equation in terms of cos x only:
\(\sin ^{2} x=\cos ^{2} x \Leftrightarrow 3\left(1-\cos ^{2} x\right)=\cos ^{2} x \Leftrightarrow 4 \cos ^{2} x=3 \Leftrightarrow \cos ^{2} x=3 / 4 \Leftrightarrow \cos x=\pm \sqrt{3} / 2\)
Solutions: x = 30° or x = 150°
– By expressing the equation in terms of tan x only:
\(\sin ^{2} x=\cos ^{2} x \Leftrightarrow \tan ^{2} x=\frac{1}{3} \Rightarrow \tan x=\pm \frac{1}{\sqrt{3}} \Rightarrow x=30^{\circ} \text { or } x=150^{\circ}\)

Question

[Maximum mark: 11] [with GDC]
(a) Consider the equation \(x^{2}+k x+1=0\). For what values of k does this equation have two equal roots? [3]
Let f be the function \(f(\theta)=2 \cos 2 \theta+4 \cos \theta+3, \text { for }-360^{\circ} \leq x \leq 360^{\circ}\).
(b) Show that this function may be written as \(f(\theta)=4 \cos ^{2} \theta+4 \cos \theta+1\)  [1]
(c) Consider the equation \(f(\theta)=0 \text {, for }-360^{\circ} \leq x \leq 360^{\circ}\).
(i) How many distinct values of cosθ satisfy this equation?
(ii) Find all values of θ which satisfy this equation. [5]
(d) Given that \(f(\theta)=c\) is satisfied by only three values of θ, find the value of c.
[hint: look at the graph of the function f(θ) ]   [2]

Answer/Explanation

Ans:
(a) \(\Delta=0 \Leftrightarrow k^{2}-4 \times 4 \times 1=0 \Leftrightarrow k^{2}=16 \Leftrightarrow k=4, k=-4\)
(b) using cos \(2 \theta=2 \cos ^{2} \theta-1\)
\(f(\theta)=4 \cos ^{2} \theta+4 \cos \theta+1\)
(c) (i) 1
(ii) METHOD 1
Solve for cos θ
\(\cos \theta=-\frac{1}{2}\)
θ = 240, 120, – 240, -120
METHOD 2
Directly by GDC: θ = 240, 120, -240, -120
(d) Using graph, c = 9

Question

[Maximum mark: 10] [with GDC]
The diagram below shows a plan for a window in the shape of a trapezium.

Three sides of the window are 2 m long. The angle between the sloping sides of the window and the base is θ, where \(0<\theta<\frac{\pi}{2}\).
(a) Show that the area of the window is given by \(y=4 \sin \theta+2 \sin 2 \theta\). [5]
(b) Zoe wants a window to have an area of 5 m2. Find the two possible values of θ. [3]
(c) John wants two windows which have the same area A but different values of θ.
Find all possible values for A.
[hint: in other words, find the range y, except max/min values] [2]

Answer/Explanation

Ans:
(a) For the height h, \(\sin \theta=\frac{h}{2} \Leftrightarrow h=2 \sin \theta\)
For the base of triangle b, \(\cos \theta=\frac{b}{2} \Leftrightarrow b=2 \cos \theta\)
\(\text { Area } y=2\left(\frac{1}{2} \times 2 \cos \theta \times 2 \sin \theta\right)+2 \times 2 \sin \theta=4 \sin \theta \cos \theta+4 \sin \theta\)
y = 4 sin θ + 2 sin 2θ
(b) 4 sin θ + 2 sin 2θ = 5
θ = 0.856 (49.0º), θ = 1.25 (71.4º)
(c) By graph GDC 4 < A < 5.20 (accept 4 < A < 5.19) 

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