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]
Find an expression for \(\cos 140^\circ \) .[3]
Find an expression for \(\tan 140^\circ \) .[1]
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) \(y\) .[2]
Let the area of the rectangle be A.
Show that \(A = 18\sin 2\theta \) .[3]
(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]
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
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]
Find an expression for \(\cos 140^\circ \) .[3]
Find an expression for \(\tan 140^\circ \) .[1]
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
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]
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]
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]
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) \(\tan x=\frac{p}{\sqrt{1-p^{2}}}, \quad \cos 2 x=1-2 p^{2}\)
\(\sin 2 x=2 p \sqrt{1-p^{2}}, \quad \tan 2 x=\frac{2 p \sqrt{1-p^{2}}}{1-2 p^{2}}\)
\(\sin 4 x=4 p \sqrt{1-p^{2}}\left(1-2 p^{2}\right)\)