IB Math Analysis & Approaches Questionbank-Topic: SL 3.6 The Pythagorean identity and Relationship between trigonometric ratios 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

Let  \(f(x) = {\sin ^3}x + {\cos ^3}x\tan x,\frac{\pi }{2} < x < \pi \) .

Show that \(f(x) = \sin x\) .

[2]
a.

Let \(\sin x = \frac{2}{3}\) . Show that \(f(2x) = – \frac{{4\sqrt 5 }}{9}\) .

[5]
b.
Answer/Explanation

Markscheme

changing \(\tan x\) into \(\frac{{\sin x}}{{\cos x}}\)     A1

e.g. \({\sin ^3}x + {\cos ^3}x\frac{{\sin x}}{{\cos x}}\)

simplifying     A1

e.g \(\sin x({\sin ^2}x + {\cos ^2}x)\) , \({\sin ^3}x + \sin x – {\sin ^3}x\)

\(f(x) = \sin x\)    AG     N0

[2 marks]

a.

recognizing \(f(2x) = \sin 2x\) , seen anywhere     (A1)

evidence of using double angle identity \(\sin (2x) = 2\sin x\cos x\) , seen anywhere     (M1)

evidence of using Pythagoras with \(\sin x = \frac{2}{3}\)     M1

e.g. sketch of right triangle, \({\sin ^2}x + {\cos ^2}x = 1\)

\(\cos x = – \frac{{\sqrt 5 }}{3}\) (accept \(\frac{{\sqrt 5 }}{3}\) )     (A1)

\(f(2x) = 2\left( {\frac{2}{3}} \right)\left( { – \frac{{\sqrt 5 }}{3}} \right)\)     A1

\(f(2x) = – \frac{{4\sqrt 5 }}{9}\)     AG     N0

[5 marks]

b.

Question

Solve \(\cos 2x – 3\cos x – 3 – {\cos ^2}x = {\sin ^2}x\) , for \(0 \le x \le 2\pi \) .

Answer/Explanation

Markscheme

evidence of substituting for \(\cos 2x\)     (M1)

evidence of substituting into \({\sin ^2}x + {\cos ^2}x = 1\)     (M1)

correct equation in terms of \(\cos x\) (seen anywhere)     A1

e.g. \(2{\cos ^2}x – 1 – 3\cos x – 3 = 1\) , \(2{\cos ^2}x – 3\cos x – 5 = 0\)

evidence of appropriate approach to solve     (M1)

e.g. factorizing, quadratic formula

appropriate working     A1

e.g. \((2\cos x – 5)(\cos x + 1) = 0\) , \((2x – 5)(x + 1)\) , \(\cos x = \frac{{3 \pm \sqrt {49}}}{4}\)

correct solutions to the equation

e.g. \(\cos x = \frac{5}{2}\) , \(\cos x = – 1\) , \(x = \frac{5}{2}\) , \(x = – 1\)     (A1)

\(x = \pi \)     A1     N4

[7 marks]

Question

Let \(f(x) = {(\sin x + \cos x)^2}\) .

Show that \(f(x)\) can be expressed as \(1 + \sin 2x\) .

[2]
a.

The graph of f is shown below for \(0 \le x \le 2\pi \) .


Let \(g(x) = 1 + \cos x\) . On the same set of axes, sketch the graph of g for \(0 \le x \le 2\pi \) .

 

[2]
b.

The graph of g can be obtained from the graph of f under a horizontal stretch of scale factor p followed by a translation by the vector \(\left( {\begin{array}{*{20}{c}}
k\\
0
\end{array}} \right)\) .

Write down the value of p and a possible value of k .

[2]
c.
Answer/Explanation

Markscheme

attempt to expand     (M1)

e.g. \((\sin x + \cos x)(\sin x + \cos x)\) ; at least 3 terms

correct expansion     A1

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

\(f(x) = 1 + \sin 2x\)     AG     N0

[2 marks]

a.


     A1A1     N2

Note: Award A1 for correct sinusoidal shape with period \(2\pi \) and range \([0{\text{, }}2]\), A1 for minimum in circle.

b.

\(p = 2\) , \(k = – \frac{\pi }{2}\)     A1A1     N2

[2 marks]

c.

Question

Given that \(\sin x = \frac{3}{4}\), where \(x\) is an obtuse angle,

find the value of \(\cos x;\)

[4]
a.

find the value of \(\cos 2x.\)

[3]
b.
Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\;\;\;\), \({\sin ^2}x + {\cos ^2}x = 1\)

correct working     (A1)

eg\(\;\;\;{4^2} – {3^2},{\text{ }}{\cos ^2}x = 1 – {\left( {\frac{3}{4}} \right)^2}\)

correct calculation     (A1)

eg\(\;\;\;\frac{{\sqrt 7 }}{4},{\text{ }}{\cos ^2}x = \frac{7}{{16}}\)

\(\cos x =  – \frac{{\sqrt 7 }}{4}\)     A1     N3

[4 marks]

a.

correct substitution (accept missing minus with cos)     (A1)

eg\(\;\;\;1 – 2{\left( {\frac{3}{4}} \right)^2},{\text{ }}2{\left( { – \frac{{\sqrt 7 }}{4}} \right)^2} – 1,{\text{ }}{\left( {\frac{{\sqrt 7 }}{4}} \right)^2} – {\left( {\frac{3}{4}} \right)^2}\)

correct working     A1

eg\(\;\;\;2\left( {\frac{7}{{16}}} \right) – 1,{\text{ }}1 – \frac{{18}}{{16}},{\text{ }}\frac{7}{{16}} – \frac{9}{{16}}\)

\(\cos 2x =  – \frac{2}{{16}}\;\;\;\left( { =  – \frac{1}{8}} \right)\)     A1     N2

[3 marks]

Total [7 marks]

b.

Question

Let \(f(x) = 6x\sqrt {1 – {x^2}} \), for \( – 1 \leqslant x \leqslant 1\), and \(g(x) = \cos (x)\), for \(0 \leqslant x \leqslant \pi \).

Let \(h(x) = (f \circ g)(x)\).

Write \(h(x)\) in the form \(a\sin (bx)\), where \(a,{\text{ }}b \in \mathbb{Z}\).

[5]
a.

Hence find the range of \(h\).

[2]
b.
Answer/Explanation

Markscheme

attempt to form composite in any order     (M1)

eg\(\,\,\,\,\,\)\(f\left( {g(x)} \right),{\text{ }}\cos \left( {6x\sqrt {1 – {x^2}} } \right)\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6\cos x\sqrt {1 – {{\cos }^2}x} \)

correct application of Pythagorean identity (do not accept \({\sin ^2}x + {\cos ^2}x = 1\))     (A1)

eg\(\,\,\,\,\,\)\({\sin ^2}x = 1 – {\cos ^2}x,{\text{ }}6\cos x\sin x,{\text{ }}6\cos x \left| \sin x\right|\)

valid approach (do not accept \(2\sin x\cos x = \sin 2x\))     (M1)

eg\(\,\,\,\,\,\)\(3(2\cos x\sin x)\)

\(h(x) = 3\sin 2x\)    A1     N3

[5 marks]

a.

valid approach     (M1)

eg\(\,\,\,\,\,\)amplitude \( = 3\), sketch with max and min \(y\)-values labelled, \( – 3 < y < 3\)

correct range     A1     N2

eg\(\,\,\,\,\,\)\( – 3 \leqslant y \leqslant 3\), \([ – 3,{\text{ }}3]\) from \( – 3\) to 3

Note:     Do not award A1 for \( – 3 < y < 3\) or for “between \( – 3\) and 3”.

[2 marks]

b.

Examiners report

In part (a), nearly all candidates found the correct composite function in terms of \(\cos x\), though many did not get any further than this first step in their solution to the question. While some candidates seemed to recognize the need to use trigonometric identities, most were unsuccessful in finding the correct expression in the required form.

a.

In part (b), very few candidates were able to provide the correct range of the function.

b.

Question

Let \(\sin \theta  = \frac{{\sqrt 5 }}{3}\), where \(\theta \) is acute.

Find \(\cos \theta \).

[3]
a.

Find \(\cos 2\theta \).

[2]
b.
Answer/Explanation

Markscheme

evidence of valid approach     (M1)

eg\(\,\,\,\,\,\)right triangle, \({\cos ^2}\theta  = 1 – {\sin ^2}\theta \)

correct working     (A1)

eg\(\,\,\,\,\,\)missing side is 2, \(\sqrt {1 – {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}} \)

\(\cos \theta  = \frac{2}{3}\)     A1     N2

[3 marks]

a.

correct substitution into formula for \(\cos 2\theta \)     (A1)

eg\(\,\,\,\,\,\)\(2 \times {\left( {\frac{2}{3}} \right)^2} – 1,{\text{ }}1 – 2{\left( {\frac{{\sqrt 5 }}{3}} \right)^2},{\text{ }}{\left( {\frac{2}{3}} \right)^2} – {\left( {\frac{{\sqrt 5 }}{3}} \right)^2}\)

\(\cos 2\theta  =  – \frac{1}{9}\)    A1     N2

[2 marks]

b.

Question

Solve \({\log _2}(2\sin x) + {\log _2}(\cos x) =  – 1\), for \(2\pi  < x < \frac{{5\pi }}{2}\).

Answer/Explanation

Markscheme

correct application of \(\log a + \log b = \log ab\)     (A1)

eg\(\,\,\,\,\,\)\({\log _2}(2\sin x\cos x),{\text{ }}\log 2 + \log (\sin x) + \log (\cos x)\)

correct equation without logs     A1

eg\(\,\,\,\,\,\)\(2\sin x\cos x = {2^{ – 1}},{\text{ }}\sin x\cos x = \frac{1}{4},{\text{ }}\sin 2x = \frac{1}{2}\)

recognizing double-angle identity (seen anywhere)     A1

eg\(\,\,\,\,\,\)\(\log (\sin 2x),{\text{ }}2\sin x\cos x = \sin 2x,{\text{ }}\sin 2x = \frac{1}{2}\)

evaluating \({\sin ^{ – 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}{\text{ }}(30^\circ )\)     (A1)

correct working     A1

eg\(\,\,\,\,\,\)\(x = \frac{\pi }{{12}} + 2\pi ,{\text{ }}2x = \frac{{25\pi }}{6},{\text{ }}\frac{{29\pi }}{6},{\text{ }}750^\circ ,{\text{ }}870^\circ ,{\text{ }}x = \frac{\pi }{{12}}\)and \(x = \frac{{5\pi }}{{12}}\), one correct final answer

\(x = \frac{{25\pi }}{{12}},{\text{ }}\frac{{29\pi }}{{12}}\) (do not accept additional values)     A2     N0

[7 marks]

Question

Let \(f(x) = 15 – {x^2}\), for \(x \in \mathbb{R}\). The following diagram shows part of the graph of \(f\) and the rectangle OABC, where A is on the negative \(x\)-axis, B is on the graph of \(f\), and C is on the \(y\)-axis.

N17/5/MATME/SP1/ENG/TZ0/06

Find the \(x\)-coordinate of A that gives the maximum area of OABC.

Answer/Explanation

Markscheme

attempt to find the area of OABC     (M1)

eg\(\,\,\,\,\,\)\({\text{OA}} \times {\text{OC, }}x \times f(x),{\text{ }}f(x) \times ( – x)\)

correct expression for area in one variable     (A1)

eg\(\,\,\,\,\,\)\({\text{area}} = x(15 – {x^2}),{\text{ }}15x – {x^3},{\text{ }}{x^3} – 15x\)

valid approach to find maximum area (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)\(A’(x) = 0\)

correct derivative     A1

eg\(\,\,\,\,\,\)\(15 – 3{x^2},{\text{ }}(15 – {x^2}) + x( – 2x) = 0,{\text{ }} – 15 + 3{x^2}\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(15 = 3{x^2},{\text{ }}{x^2} = 5,{\text{ }}x = \sqrt 5 \)

\(x =  – \sqrt 5 {\text{ }}\left( {{\text{accept A}}\left( { – \sqrt 5 ,{\text{ }}0} \right)} \right)\)     A2     N3

[7 marks]

Examiners report

[N/A]

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

Given that \(\cos A = \frac{1}{3}\) and \(0 \le A \le \frac{\pi }{2}\) , find \(\cos 2A\) .

[3]
a.

Given that \(\sin B = \frac{2}{3}\)  and \(\frac{\pi }{2} \le B \le \pi \) , find \(\cos B\) .

[3]
b.
Answer/Explanation

Markscheme

evidence of choosing the formula \(\cos 2A = 2{\cos ^2}A – 1\)     (M1)

Note: If they choose another correct formula, do not award the M1 unless there is evidence of finding \({\sin ^2}A = 1 – \frac{1}{9}\)

correct substitution     A1

 e.g.\(\cos 2A = {\left( {\frac{1}{3}} \right)^2} – \frac{8}{9}\) , \(\cos 2A = 2 \times {\left( {\frac{1}{3}} \right)^2} – 1\)

 \(\cos 2A = – \frac{7}{9}\)     A1     N2

[3 marks]

a.

METHOD 1

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

e.g. \({\left( {\frac{2}{3}} \right)^2} + {\cos ^2}B = 1\) , \(\sqrt {\frac{5}{9}} \) (seen anywhere),

\(\cos B = \pm \sqrt {\frac{5}{9}} \) \(\left( { = \pm \frac{{\sqrt 5 }}{3}} \right)\)     (A1)

 \(\cos B = – \sqrt {\frac{5}{9}} \) \(\left( { = – \frac{{\sqrt 5 }}{3}} \right)\)     A1     N2

METHOD 2

diagram     M1


for finding third side equals \(\sqrt 5 \)     (A1)

\(\cos B = – \frac{{\sqrt 5 }}{3}\)     A1     N2

[3 marks]

b.

Question

Let  \(f(x) = {\sin ^3}x + {\cos ^3}x\tan x,\frac{\pi }{2} < x < \pi \) .

Show that \(f(x) = \sin x\) .

[2]
a.

Let \(\sin x = \frac{2}{3}\) . Show that \(f(2x) = – \frac{{4\sqrt 5 }}{9}\) .

[5]
b.
Answer/Explanation

Markscheme

changing \(\tan x\) into \(\frac{{\sin x}}{{\cos x}}\)     A1

e.g. \({\sin ^3}x + {\cos ^3}x\frac{{\sin x}}{{\cos x}}\)

simplifying     A1

e.g \(\sin x({\sin ^2}x + {\cos ^2}x)\) , \({\sin ^3}x + \sin x – {\sin ^3}x\)

\(f(x) = \sin x\)    AG     N0

[2 marks]

a.

recognizing \(f(2x) = \sin 2x\) , seen anywhere     (A1)

evidence of using double angle identity \(\sin (2x) = 2\sin x\cos x\) , seen anywhere     (M1)

evidence of using Pythagoras with \(\sin x = \frac{2}{3}\)     M1

e.g. sketch of right triangle, \({\sin ^2}x + {\cos ^2}x = 1\)

\(\cos x = – \frac{{\sqrt 5 }}{3}\) (accept \(\frac{{\sqrt 5 }}{3}\) )     (A1)

\(f(2x) = 2\left( {\frac{2}{3}} \right)\left( { – \frac{{\sqrt 5 }}{3}} \right)\)     A1

\(f(2x) = – \frac{{4\sqrt 5 }}{9}\)     AG     N0

[5 marks]

b.

Question

The vertices of the triangle PQR are defined by the position vectors

\(\overrightarrow {{\rm{OP}}}  = \left( {\begin{array}{*{20}{c}}
4\\
{ – 3}\\
1
\end{array}} \right)\) , \(\overrightarrow {{\rm{OQ}}}  = \left( {\begin{array}{*{20}{c}}
3\\
{ – 1}\\
2
\end{array}} \right)\) and \(\overrightarrow {{\rm{OR}}}  = \left( {\begin{array}{*{20}{c}}
6\\
{ – 1}\\
5
\end{array}} \right)\) .

Find

(i)     \(\overrightarrow {{\rm{PQ}}} \) ;

(ii)    \(\overrightarrow {{\rm{PR}}} \) .

[3]
a.

Show that \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) .

[7]
b.

(i)     Find \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}}\) .

(ii)    Hence, find the area of triangle PQR, giving your answer in the form \(a\sqrt 3 \) .

[6]
c.
Answer/Explanation

Markscheme

(i) evidence of approach     (M1)

e.g. \(\overrightarrow {{\rm{PQ}}}  = \overrightarrow {{\rm{PO}}}  + \overrightarrow {{\rm{OQ}}} \) , \({\rm{Q}} – {\rm{P}}\)

\(\overrightarrow {{\rm{PQ}}}  = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
1
\end{array}} \right)\)     A1     N2

(ii) \(\overrightarrow {{\rm{PR}}}  = \left( {\begin{array}{*{20}{c}}
2\\
2\\
4
\end{array}} \right)\)     A1     N1

[3 marks]

a.

METHOD 1

choosing correct vectors \(\overrightarrow {{\rm{PQ}}} \) and \(\overrightarrow {{\rm{PR}}} \)    (A1)(A1)

finding \(\overrightarrow {{\rm{PQ}}}  \bullet \overrightarrow {{\rm{PR}}} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right|\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right|\)    (A1) (A1)(A1)

\(\overrightarrow {{\rm{PQ}}}  \bullet \overrightarrow {{\rm{PR}}}  = – 2 + 4 + 4( = 6)\)

\(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt {{{( – 1)}^2} + {2^2} + {1^2}} \) \(\left( { = \sqrt 6 } \right)\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {{2^2} + {2^2} + {4^2}} \) \(\left( { = \sqrt {24} } \right)\)

substituting into formula for angle between two vectors     M1

e.g. \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{6}{{\sqrt 6  \times \sqrt {24} }}\)

simplifying to expression clearly leading to \(\frac{1}{2}\)     A1

e.g. \(\frac{6}{{\sqrt 6  \times 2\sqrt 6 }}\) , \(\frac{6}{{\sqrt {144} }}\) , \(\frac{6}{{12}}\)

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\)     AG     N0

METHOD 2

evidence of choosing cosine rule (seen anywhere)     (M1)

\(\overrightarrow {{\rm{QR}}}  = \left( {\begin{array}{*{20}{c}}
3\\
0\\
3
\end{array}} \right)\)     A1

\(\left| {\overrightarrow {{\rm{QR}}} } \right| = \sqrt {18} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt 6 \) and \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {24} \)     (A1)(A1)(A1)

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt {24} } \right)}^2} – {{\left( {\sqrt {18} } \right)}^2}}}{{2\sqrt 6  \times \sqrt {24} }}\)     A1

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{6 + 24 – 18}}{{24}}\) \(\left( { = \frac{{12}}{{24}}} \right)\)     A1

\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\)      AG     N0

[7 marks]

b.

(i) METHOD 1

evidence of appropriate approach     (M1)

e.g. using \({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q + co}}{{\rm{s}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q}} = 1\) , diagram

substituting correctly     (A1)

e.g. \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {1 – {{\left( {\frac{1}{2}} \right)}^2}} \)

\({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {\frac{3}{4}} \) \(\left( { = \frac{{\sqrt 3 }}{2}} \right)\)     A1     N3

METHOD 2

since \(\cos \widehat {\rm{P}} = \frac{1}{2}\) , \(\widehat {\rm{P}} = 60^\circ \)     (A1)

evidence of approach

e.g. drawing a right triangle, finding the missing side     (A1)

\(\sin \widehat {\rm{P}} = \frac{{\sqrt 3 }}{2}\)     A1     N3

(ii) evidence of appropriate approach      (M1)

e.g. attempt to substitute into \(\frac{1}{2}ab\sin C\)

correct substitution

e.g. area \( = \frac{1}{2}\sqrt 6  \times \sqrt {24}  \times \frac{{\sqrt 3 }}{2}\)     A1

area \( = 3\sqrt 3 \)     A1     N2

[6 marks]

c.

Question

Let \(\sin \theta  = \frac{2}{{\sqrt {13} }}\) , where \(\frac{\pi }{2} < \theta  < \pi \) .

Find \(\cos \theta \) .

[3]
a.

Find \(\tan 2\theta \) .

[5]
b.
Answer/Explanation

Markscheme

METHOD 1

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

correct working     (A1)

e.g. \({\cos ^2}\theta  = \frac{9}{{13}}\) , \(\cos \theta  =  \pm \frac{3}{{\sqrt {13} }}\) , \(\cos \theta  = \sqrt {\frac{9}{{13}}} \)

\(\cos \theta  = – \frac{3}{{\sqrt {13} }}\)    A1     N2

Note: If no working shown, award N1 for \(\frac{3}{{\sqrt {13} }}\) .

METHOD 2

approach involving Pythagoras’ theorem     (M1)

e.g. \({2^2} + {x^2} = 13\) ,


finding third side equals 3     (A1)

\(\cos \theta  = – \frac{3}{{\sqrt {13} }}\)     A1     N2

Note: If no working shown, award N1 for \(\frac{3}{{\sqrt {13} }}\) .

[3 marks]

a.

correct substitution into \(\sin 2\theta \) (seen anywhere)     (A1)

e.g. \(2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { – \frac{3}{{\sqrt {13} }}} \right)\)

correct substitution into \(\cos 2\theta \) (seen anywhere)     (A1)

e.g. \({\left( { – \frac{3}{{\sqrt {13} }}} \right)^2} – {\left( {\frac{2}{{\sqrt {13} }}} \right)^2}\) , \(2{\left( { – \frac{3}{{\sqrt {13} }}} \right)^2} – 1\) , \(1 – 2{\left( {\frac{2}{{\sqrt {13} }}} \right)^2}\)

valid attempt to find \(\tan 2\theta \)     (M1)

e.g. \(\frac{{2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { – \frac{3}{{\sqrt {13} }}} \right)}}{{{{\left( { – \frac{3}{{\sqrt {13} }}} \right)}^2} – {{\left( {\frac{2}{{\sqrt {13} }}} \right)}^2}}}\) , \(\frac{{2\left( { – \frac{2}{3}} \right)}}{{1 – {{\left( { – \frac{2}{3}} \right)}^2}}}\)

correct working     A1

e.g. \(\frac{{\frac{{(2)(2)( – 3)}}{{13}}}}{{\frac{9}{{13}} – \frac{4}{{13}}}}\) , \(\frac{{ – \frac{{12}}{{{{\left( {\sqrt {13} } \right)}^2}}}}}{{\frac{{18}}{{13}} – 1}}\) , \(\frac{{ – \frac{{12}}{{13}}}}{{\frac{5}{{13}}}}\)

\(\tan 2\theta  = – \frac{{12}}{5}\)     A1     N4

Note: If students find answers for \(\cos \theta \) which are not in the range \([ – 1{\text{, }}1]\), award full FT in (b) for correct FT working shown.

[5 marks]

b.

Question

Let \(f(x) = {(\sin x + \cos x)^2}\) .

Show that \(f(x)\) can be expressed as \(1 + \sin 2x\) .

[2]
a.

The graph of f is shown below for \(0 \le x \le 2\pi \) .


Let \(g(x) = 1 + \cos x\) . On the same set of axes, sketch the graph of g for \(0 \le x \le 2\pi \) .

 

[2]
b.

The graph of g can be obtained from the graph of f under a horizontal stretch of scale factor p followed by a translation by the vector \(\left( {\begin{array}{*{20}{c}}
k\\
0
\end{array}} \right)\) .

Write down the value of p and a possible value of k .

[2]
c.
Answer/Explanation

Markscheme

attempt to expand     (M1)

e.g. \((\sin x + \cos x)(\sin x + \cos x)\) ; at least 3 terms

correct expansion     A1

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

\(f(x) = 1 + \sin 2x\)     AG     N0

[2 marks]

a.


     A1A1     N2

Note: Award A1 for correct sinusoidal shape with period \(2\pi \) and range \([0{\text{, }}2]\), A1 for minimum in circle.

b.

\(p = 2\) , \(k = – \frac{\pi }{2}\)     A1A1     N2

[2 marks]

c.

Question

Let \(\sin {100^ \circ } = m\). Find an expression for \(\cos {100^ \circ }\) in terms of m.

[3]
a.

Let \(\sin {100^ \circ } = m\) . Find an expression for \(\tan {100^ \circ }\) in terms of m.

[1]
b.

Let \(\sin {100^ \circ } = m\). Find an expression for \(\sin {200^ \circ }\) in terms of m.

[2]
c.
Answer/Explanation

Markscheme

Note: All answers must be given in terms of m. If a candidate makes an error that means there is no m in their answer, do not award the final A1FT mark.

METHOD 1 

valid approach involving Pythagoras     (M1)

e.g. \({\sin ^2}x + {\cos ^2}x = 1\) , labelled diagram 

correct working (may be on diagram)     (A1)

e.g. \({m^2} + {(\cos 100)^2} = 1\) , \(\sqrt {1 – {m^2}} \)

\(\cos 100 = – \sqrt {1 – {m^2}} \)    A1     N2

[3 marks]

METHOD 2

valid approach involving tan identity     (M1)

e.g. \(\tan  = \frac{{\sin }}{{\cos }}\)

correct working     (A1)

e.g. \(\cos 100 = \frac{{\sin 100}}{{\tan 100}}\)

\(\cos 100 = \frac{m}{{\tan 100}}\)     A1     N2

[3 marks]

a.

METHOD 1

\(\tan 100 = – \frac{m}{{\sqrt {1 – {m^2}} }}\) (accept \(\frac{m}{{ – \sqrt {1 – {m^2}} }}\))     A1     N1

[1 mark] 

METHOD 2

\(\tan 100 = \frac{m}{{\cos 100}}\)     A1     N1

[1 mark]

b.

METHOD 1

valid approach involving double angle formula     (M1)

e.g. \(\sin 2\theta = 2\sin \theta cos\theta \)

\(\sin 200 = – 2m\sqrt {1 – {m^2}} \)  (accept \(2m\left( { – \sqrt {1 – {m^2}} } \right)\))     A1     N2

Note: If candidates find \(\cos 100 = \sqrt {1 – {m^2}} \) , award full FT in parts (b) and (c), even though the values may not have appropriate signs for the angles.

[2 marks]

METHOD 2

valid approach involving double angle formula     (M1)

e.g. \(\sin 2\theta  = 2\sin \theta \cos \theta \) , \(2m \times \frac{m}{{\tan 100}}\)

\(\sin 200 = \frac{{2{m^2}}}{{\tan 100}}( = 2m\cos 100)\)     A1     N2

[2 marks]

c.

Question

The following diagram shows a right-angled triangle, \(\rm{ABC}\), where \(\sin \rm{A} = \frac{5}{{13}}\).

Show that \(\cos A = \frac{{12}}{{13}}\).

[2]
a.

Find \(\cos 2A\).

[3]
b.
Answer/Explanation

Markscheme

METHOD 1

approach involving Pythagoras’ theorem     (M1)

eg     \({5^2} + {x^2} = {13^2}\), labelling correct sides on triangle

finding third side is 12 (may be seen on diagram)     A1

\(\cos A = \frac{{12}}{{13}}\)     AG     N0

METHOD 2

approach involving \({\sin ^2}\theta  + {\cos ^2}\theta  = 1\)     (M1)

eg     \({\left( {\frac{5}{{13}}} \right)^2} + {\cos ^2}\theta  = 1,{\text{ }}{x^2} + \frac{{25}}{{169}} = 1\)

correct working     A1

eg     \({\cos ^2}\theta  = \frac{{144}}{{169}}\)

\(\cos A = \frac{{12}}{{13}}\)     AG     N0

[2 marks]

a.

correct substitution into \(\cos 2\theta \)     (A1)

eg     \(1 – 2{\left( {\frac{5}{{13}}} \right)^2},{\text{ }}2{\left( {\frac{{12}}{{13}}} \right)^2} – 1,{\text{ }}{\left( {\frac{{12}}{{13}}} \right)^2} – {\left( {\frac{5}{{13}}} \right)^2}\)

correct working     (A1)

eg     \(1 – \frac{{50}}{{169}},{\text{ }}\frac{{288}}{{169}} – 1,{\text{ }}\frac{{144}}{{169}} – \frac{{25}}{{169}}\)

\(\cos 2A = \frac{{119}}{{169}}\)     A1     N2

[3 marks]

b.

Leave a Reply