# CIE A level -Pure Mathematics 1 : Topic :  1.5 Trigonometry: use the identities : Exam Questions Paper 1

### Question

(i)Prove the identity $$\frac{\sin \Theta }{1-\cos \Theta }-\frac{1}{\sin \Theta }=\frac{1}{\tan \Theta }$$.

(ii) Hence solve the equation $$\frac{\sin \Theta }{1-\cos \Theta }-\frac{1}{\sin \Theta }=4\tan \Theta$$ for $$0^{\circ}< \Theta < 180^{\circ}$$.

▶️Answer/Explanation

(i) Solvung LHS
$$\frac{\sin \Theta}{1 – \cos \Theta} – \frac{1}{\sin \Theta}$$
$$= \frac{\sin^2 \Theta – (1 – \cos \Theta)}{(1 – \cos \Theta)\sin \Theta}$$
$$= \frac{\sin^2 \Theta – 1 + \cos \Theta}{(1 – \cos \Theta)\sin \Theta}$$
$$= \frac{1 – \cos^2 \Theta – 1 + \cos \Theta}{(1 – \cos \Theta)\sin \Theta}$$
$$= \frac{\cos \Theta – \cos^2 \Theta}{(1 – \cos \Theta)\sin \Theta}$$
$$= \frac{\cos \Theta(1 – \cos \Theta)}{(1 – \cos \Theta)\sin \Theta}$$
$$= \frac{\cos \Theta}{\sin \Theta}$$
$$= \frac{1}{\tan \Theta}$$
(ii) $$\frac{1}{\tan \Theta} = 4 \tan \Theta$$
$$1 = 4 \tan^2 \Theta$$
$$\tan^2 \Theta = \frac{1}{4}$$
$$\tan \Theta = \pm\frac{1}{2}$$
$$\Theta =26.6^{\circ}, 153.4^{\circ}$$

Question

(a)The diagram shows part of the graph of $$y=a+b\sin x$$.Find the values of constant a and b.

(b)(i)Show that the equation $$\left ( \sin \Theta +2\cos \Theta \right )\left ( 1+\sin \Theta -\cos \Theta \right )=\sin \Theta \left ( 1+\cos \Theta \right )$$

may be expressed as $$3\cos ^{2}\Theta -2\cos \Theta -1=0$$.

(c)Hence solve the equation $$\left ( \sin \Theta +2\cos \Theta \right )\left ( 1+\sin \Theta-\cos \Theta \right )=\sin \Theta \left ( 1+\cos \Theta \right )$$ for $$-180^{\circ}\leq \Theta \leq 180^{\circ}$$

Answer/Explanation

(i)a=-2,b=3

(b)(ii)$$s+s^{2}-sc+2c+2sc=s+sc\rightarrow s^{2}-2c^{2}+2c=0$$

$$1-\cos ^{2}\Theta -2\cos ^{2}\Theta +2\cos \Theta =0$$

$$3\cos ^{2}\Theta -2\cos \Theta -1=0$$

(ii)$$\cos \Theta = 1$$or $$-\frac{1}{3}$$

$$\Theta =0^{\circ}$$ or $$109.5^{\circ}$$ or $$-109.5^{\circ}$$

Question

(i) Show that the equation $$\frac{4\cos \Theta }{\tan \Theta }+15=0$$ can be expressed as $$4\sin ^{2}\Theta -15\sin \Theta -4=0$$.

(ii) Hence solve the equation $$\frac{4\cos \Theta }{\tan \Theta }+15=0$$ for $$0^{\circ}\leq \Theta \leq 360^{\circ}$$.

Answer/Explanation

(i)$$4\cos ^{2}\Theta +15\sin \Theta =0$$

$$4(1-s^{2})+15s=0\rightarrow 4\sin ^{2}\Theta -15\sin \Theta -4=0$$

(ii)$$\sin \Theta =-\frac{1}{4}$$

$$\Theta$$=194.5 or 345.5

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