# CIE A level -Pure Mathematics 1 : Topic :  1.5 Trigonometry: inverse trigonometric relations : Exam Questions Paper 1

Question

(i)Show that $$\frac{\sin \Theta }{\sin \Theta +\cos \Theta }+\frac{\cos \Theta }{\sin \Theta -\cos \Theta }=\frac{1}{\sin ^{2}\Theta -\cos ^{2}\Theta }$$.

(ii)Hence solve the equation $$\frac{\sin \Theta }{\sin \Theta +\cos \Theta }+\frac{\cos \Theta }{\sin \Theta -\cos \Theta }=3$$ for $$0^{\circ}\leq \Theta \leq 360^{\circ}$$

(i)Solving LHS
$$\frac{\sin \Theta}{\sin \Theta + \cos \Theta} + \frac{\cos \Theta}{\sin \Theta – \cos \Theta}$$
$$= \frac{\sin \Theta(\sin \Theta – \cos \Theta) + \cos \Theta(\sin \Theta + \cos \Theta)}{(\sin \Theta + \cos \Theta)(\sin \Theta – \cos \Theta)}$$
$$= \frac{\sin^2 \Theta – \sin \Theta \cos \Theta + \sin \Theta \cos \Theta + \cos^2 \Theta}{\sin^2 \Theta – \cos^2 \Theta}$$
$$= \frac{\sin^2 \Theta + \cos^2 \Theta}{\sin^2 \Theta – \cos^2 \Theta}$$
$$= \frac{1}{\sin^2 \Theta – \cos^2 \Theta}$$
(ii) $$\frac{1}{\sin^2 \Theta – \cos^2 \Theta} = 3$$
$$\sin^2 \Theta – \cos^2 \Theta = \frac{1}{3}$$
$$\sin^2 \Theta – (1 – \sin^2 \Theta) = \frac{1}{3} \] (\(\cos^2 \Theta = 1 – \sin^2 \Theta$$)
$$2\sin^2 \Theta – 1 = \frac{1}{3}$$
$$2\sin^2 \Theta = \frac{4}{3}$$
$$\sin^2 \Theta = \frac{2}{3}$$
$$\sin \Theta = \pm\sqrt{\frac{2}{3}}$$
$\Theta=54.7^{\circ}, 125.3^{\circ}, 234.7^{\circ}, 305.3^{\circ}$

### Question

(i) Show that $$\cos ^{4}x=1-2\sin ^{2}x+\sin ^{4}x.$$

(ii)Hence,or otherwise,solve the equation $$8\sin ^{4}x+\cos ^{4}x=2\cos ^{2}x$$  for $$0^{\circ}\leq x\leq 360^{\circ}$$

(i) $$\cos^4 x = (\cos^2 x)^2$$
$$= (1 – \sin^2 x)^2$$ \
$$= 1 – 2\sin^2 x + \sin^4 x$$
Hence,proved.
(ii) $$8 \sin^4 x + (1 – 2 \sin^2 x + \sin^4 x) = 2 \cos^2 x$$
$$8 \sin^4 x + 1 – 2 \sin^2 x + \sin^4 x = 2(1 – \sin^2 x)$$ (again using $$\cos^2 x = 1 – \sin^2 x$$)
$$9 \sin^4 x – 2 \sin^2 x + 1 = 2 – 2 \sin^2 x$$
$$9 \sin^4 x – 2 \sin^2 x + 1 – 2 + 2 \sin^2 x = 0$$
$$9 \sin^4 x – 1 = 0$$
$$\sin^4 x = \frac{1}{9}$$
$$\sin^2 x = \pm\sqrt{\frac{1}{9}}$$
$$\sin^2 x = \pm\frac{1}{3}$$
As, $$\sin^2 x$$ cannot be negative, we will only consider the positive solution:
$$\sin^2 x = \frac{1}{3}$$
$$\sin x = \pm\sqrt{\frac{1}{3}}$$
$$\sin x = \pm\frac{1}{\sqrt{3}}$$
For $$\sin x = \frac{1}{\sqrt{3}}$$:
$$x = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$$ (First Quadrant)
$$x = 180^\circ – \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$$ (Second Quadrant)
For $$\sin x = -\frac{1}{\sqrt{3}}$$:
$$x = 180^\circ + \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$$ (Third Quadrant)
$$x = 360^\circ – \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)$$ (Fourth Quadrant)
\Theta will have values $35.3^{\circ}$,$144.7^{\circ}, 215.3^{\circ}, 324.7^{\circ}$

Question

Solve the equation $$\sin ^{-1}\left ( 4x^{4}+x^{2} \right )=\frac{1}{6}\pi$$.

$$4x^4 + x^2 = \sin\left(\frac{\pi}{6}\right)$$
$$4x^4 + x^2 = 0.5$$
$$4y^2 + y – 0.5 = 0$$
$$y = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
where $$a = 4$$, $$b = 1$$, and $$c = -0.5$$.
$$y = \frac{-1 \pm \sqrt{1 – 4 \cdot 4 \cdot -0.5}}{2 \cdot 4}$$
$$y = \frac{-1 \pm \sqrt{1 + 8}}{8}$$
$$y = \frac{-1 \pm \sqrt{9}}{8}$$
$$y = \frac{-1 \pm 3}{8}$$
$$y = \frac{-1 + 3}{8} = \frac{2}{8} = \frac{1}{4}$$
$$y = \frac{-1 – 3}{8} = \frac{-4}{8} = -\frac{1}{2}$$
However, $$y = x^2$$ can not be negative.Thus, $$x^2 = \frac{1}{4}$$,
$$x = \pm\sqrt{\frac{1}{4}}$$
$$x = \pm\frac{1}{2}$$
So, the solutions to the original equation are $$x = \frac{1}{2}$$ and $$x = -\frac{1}{2}$$.

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