# CIE A level -Pure Mathematics 1 : Topic :  1.5 Trigonometry: exact values of the sine, cosine and tangent : Exam Questions Paper 1

### Question

Given that $$\Theta$$ is an obtuse angle measured in radians and that $$\sin \Theta =k$$,find ,in terms of k ,an expression for

(i)$$\cos \Theta$$

(ii)$$\tan \Theta$$

(iii)$$\sin \left ( \Theta +\pi \right )$$

(i)$$\sin^2 \Theta + \cos^2 \Theta = 1$$
$$k^2 + \cos^2 \Theta = 1$$
$$\cos^2 \Theta = 1 – k^2$$
Since $$\Theta$$ is in the second quadrant, $$\cos \Theta$$ is negative. Therefore:
$$\cos \Theta = -\sqrt{1 – k^2}$$
(ii) $$\tan \Theta = \frac{\sin \Theta}{\cos \Theta} = \frac{k}{-\sqrt{1 – k^2}}$$
(iii) $$\sin (\Theta + \pi) = \sin \Theta \cos \pi + \cos \Theta \sin \pi$$ ($$\sin (A + B) = \sin A \cos B + \cos A \sin B$$)
$$\sin (\Theta + \pi) = k(-1) + 0$$
$$\sin (\Theta + \pi) = -k$$

### Question

(i) Prove the identity $$\left ( \frac{1}{cos}-tanx \right )^{2}\equiv \frac{1-sinx}{1+sinx}$$

(ii) Hence solve the equation $$\left ( \frac{1}{cosx}-tan2x \right )^{2}=\frac{1}{3}$$ for0 ≤ x ≤ 0.

(i)Solving LHS
$$\left(\frac{1}{\cos x} – \tan x\right)^2$$
$$= \left(\frac{1}{\cos x} – \frac{\sin x}{\cos x}\right)^2$$ (since $$\tan x = \frac{\sin x}{\cos x}$$)
$$= \left(\frac{1 – \sin x}{\cos x}\right)^2$$
$$= \frac{(1 – \sin x)^2}{\cos^2 x}$$
$$= \frac{1 – 2\sin x + \sin^2 x}{\cos^2 x}$$
$$= \frac{1 – 2\sin x + \sin^2 x}{1 – \sin^2 x}$$
$$= \frac{(1 – \sin x)^2}{(1 + \sin x)(1 – \sin x)}$$ (factorizing the denominator)
$$= \frac{1 – \sin x}{1 + \sin x}$$
(ii)Substituting $$\tan x$$ with $$\tan 2x$$ in the identity:
$$\left(\frac{1}{\cos x} – \tan 2x\right)^2 = \frac{1 – \sin x}{1 + \sin x}$$
$$\frac{1 – \sin x}{1 + \sin x} = \frac{1}{3}$$
$$3(1 – \sin x) = 1 + \sin x$$
$$3 – 3\sin x = 1 + \sin x$$
$$4\sin x = 2$$
$$\sin x = \frac{1}{2}$$
In the given range $$0 \leq x \leq 2\pi$$ where $$\sin x = \frac{1}{2}$$.
The solutions are
$$x = \frac{\pi}{6}, \frac{5\pi}{6}$$

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