Question
Given that \(\cos A = \frac{1}{3}\) and \(0 \le A \le \frac{\pi }{2}\) , find \(\cos 2A\) .
Given that \(\sin B = \frac{2}{3}\) and \(\frac{\pi }{2} \le B \le \pi \) , find \(\cos B\) .
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]
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]
Question
Let \(f(x) = {\sin ^3}x + {\cos ^3}x\tan x,\frac{\pi }{2} < x < \pi \) .
Show that \(f(x) = \sin x\) .
Let \(\sin x = \frac{2}{3}\) . Show that \(f(2x) = – \frac{{4\sqrt 5 }}{9}\) .
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]
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]
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
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
The straight line with equation \(y = \frac{3}{4}x\) makes an acute angle \(\theta \) with the x-axis.
Write down the value of \(\tan \theta \) .
Find the value of
(i) \(\sin 2\theta \) ;
(ii) \(\cos 2\theta \) .
Answer/Explanation
Markscheme
\(\tan \theta = \frac{3}{4}\) (do not accept \(\frac{3}{4}x\) ) A1 N1
[1 mark]
(i) \(\sin \theta = \frac{3}{5}\) , \(\cos \theta = \frac{4}{5}\) (A1)(A1)
correct substitution A1
e.g. \(\sin 2\theta = 2\left( {\frac{3}{5}} \right)\left( {\frac{4}{5}} \right)\)
\(\sin 2\theta = \frac{{24}}{{25}}\) A1 N3
(ii) correct substitution A1
e.g. \(\cos 2\theta = 1 – 2{\left( {\frac{3}{5}} \right)^2}\) , \({\left( {\frac{4}{5}} \right)^2} – {\left( {\frac{3}{5}} \right)^2}\)
\(\cos 2\theta = \frac{7}{{25}}\) A1 N1
[6 marks]
Question
Let \(f(x) = \cos 2x\) and \(g(x) = 2{x^2} – 1\) .
Find \(f\left( {\frac{\pi }{2}} \right)\) .
Find \((g \circ f)\left( {\frac{\pi }{2}} \right)\) .
Given that \((g \circ f)(x)\) can be written as \(\cos (kx)\) , find the value of k, \(k \in \mathbb{Z}\) .
Answer/Explanation
Markscheme
\(f\left( {\frac{\pi }{2}} \right) = \cos \pi \) (A1)
\( = – 1\) A1 N2
[2 marks]
\((g \circ f)\left( {\frac{\pi }{2}} \right) = g( – 1)\) \(( = 2{( – 1)^2} – 1)\) (A1)
\(= 1\) A1 N2
[2 marks]
\((g \circ f)(x) = 2{(\cos (2x))^2} – 1\) \(( = 2{\cos ^2}(2x) – 1)\) 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]
Question
Show that \(4 – \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin \theta + 3\) .
Hence, solve the equation \(4 – \cos 2\theta + 5\sin \theta = 0\) for \(0 \le \theta \le 2\pi \) .
Answer/Explanation
Markscheme
attempt to substitute \(1 – 2{\sin ^2}\theta \) for \(\cos 2\theta \) (M1)
correct substitution A1
e.g. \(4 – (1 – 2{\sin ^2}\theta ) + 5\sin\theta \)
\(4 – \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin\theta + 3\) AG N0
[2 marks]
evidence of appropriate approach to solve (M1)
e.g. factorizing, quadratic formula
correct working A1
e.g. \((2\sin \theta + 3)(\sin \theta + 1)\) , \((2x + 3)(x + 1) = 0\) , \(\sin x = \frac{{ – 5 \pm \sqrt 1 }}{4}\)
correct solution \(\sin \theta = – 1\) (do not penalise for including \(\sin \theta = – \frac{3}{2}\) (A1)
\(\theta = \frac{{3\pi }}{2}\) A2 N3
[5 marks]
Question
Let \(\sin \theta = \frac{2}{{\sqrt {13} }}\) , where \(\frac{\pi }{2} < \theta < \pi \) .
Find \(\cos \theta \) .
Find \(\tan 2\theta \) .
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]
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]
Examiners report
While the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle in the second quadrant will have a negative value.
In part (b), many candidates incorrectly tried to calculate \(\tan 2\theta \) as \(2 \times \tan \theta \) , rather than using the double-angle identities.
Question
The following diagram shows a right-angled triangle, \(\rm{ABC}\), where \(\sin \rm{A} = \frac{5}{{13}}\).
Show that \(\cos A = \frac{{12}}{{13}}\).
Find \(\cos 2A\).
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]
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]
Question
Given that \(\sin x = \frac{3}{4}\), where \(x\) is an obtuse angle,
find the value of \(\cos x;\)
find the value of \(\cos 2x.\)
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]
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]
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 \(a,{\text{ }}b \in \mathbb{Z}\).
Hence find the range of \(h\).
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]
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]
Question
Let \(\sin \theta = \frac{{\sqrt 5 }}{3}\), where \(\theta \) is acute.
Find \(\cos \theta \).
Find \(\cos 2\theta \).
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]
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]
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.
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]
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]
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 .
Hence or otherwise, solve the equation \(6\sin x\cos x = \frac{3}{2}\) , for \(\frac{\pi }{4} \le x \le \frac{\pi }{2}\) .
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]
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]