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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) \(\cos 70^\circ \) .
Find an expression for \(\cos 140^\circ \) .
Find an expression for \(\tan 140^\circ \) .
Answer/Explanation
Markscheme
(i) \(\sin 140^\circ = p\) A1 N1
(ii) \(\cos 70^\circ = – q\) A1 N1
[2 marks]
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]
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]
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) \(y\) .
Let the area of the rectangle be A.
Show that \(A = 18\sin 2\theta \) .
(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.
Answer/Explanation
Markscheme
(i) \(x = 3\cos \theta \) A1 N1
(ii) \(y = 3\sin \theta \) A1 N1
[2 marks]
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]
(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]
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 p• p + p•q + p•r
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.
Find the possible values of r.
Show that the sum of the infinite sequence is \(\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}\).
Find the values of θ which give the greatest value of the sum.
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]
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]
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]
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]