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)$$\frac{\sin \Theta\left ( \sin \Theta -\cos \Theta \right )+\cos \Theta \left ( \sin \Theta +\cos \Theta \right ) }{\left ( \sin \Theta +\cos \Theta \right )\left ( \sin \Theta -\cos \Theta \right )}$$

$$\frac{\sin^{2}\Theta -\sin \Theta \cos \Theta +\sin \Theta \cos \Theta +\cos ^{2}\Theta }{\sin ^{2}\Theta -\cos ^{2}\Theta }$$

$$\frac{1}{\sin ^{2}\Theta -\cos ^{2}\Theta }$$

(ii)$$\sin ^{2}\Theta -\left ( 1-\sin ^{2} \Theta \right )=\frac{1}{3}$$ or $$1-\cos ^{2}\Theta -\cos ^{2}\Theta =\frac{1}{3}$$

or $$3\left ( \sin ^{2} \Theta -\cos ^{2}\Theta \right )=\cos ^{2}\Theta +\sin ^{2}\Theta$$

$$\sin \Theta =\pm \sqrt{\frac{2}{3}}$$ or $$\cos \Theta =\pm \sqrt{\frac{1}{3}}$$

$$\tan \Theta =\pm \sqrt{2}$$

$$\Theta =54.7^{\circ},125.3^{\circ},234.7^{\circ},305.3^{\circ}$$

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)$$\cos ^{4}x=\left ( 1-\sin ^{2}x \right )^{2}=1-2\sin ^{2}x+\sin ^{4}x$$

(ii)$$8\sin ^{4}x+1-2\sin ^{2}x+\sin ^{4}x=2\left ( 1-\sin ^{2}x \right )$$

$$9\sin ^{4}x=1$$

$$x=35.3^{\circ}$$ (or any correct solution)

Any correct second solution from $$144.7^{\circ},215.3^{\circ},324.7^{\circ}$$

The remaining 2 solutions

Question

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

$$4x^{2}+x^{2}=\frac{1}{2}$$

Solve as quadratic in $$x^{2}$$

$$x^{2}=\frac{1}{4}$$

$$x=\pm \frac{1}{2}$$

Question

Find the value of x satisfying the equation $$\sin ^{-1}\left ( x-1 \right )=\tan ^{-1}3$$.

$$\tan ^{-1}(3)=1.249$$  or $$71.565^{\circ}$$

$$\sin 1.25$$   $$\sin 71.6$$  or 0.949

x=1.95 accept

$$1+\frac{3}{\sqrt{10}}$$

#### Question.

(a) Solve the equation $$sin^{-1}(3x) = -1$$, giving the solution in an exact form.
(b) Solve, by factorising, the equation $$2 cos\Theta sin\Theta – 2 cos\Theta – sin\Theta +1 = 0$$ for $$0 \leq \Theta \leq \Pi$$.

Ans:(a)$$(3x)=-\frac{\sqrt{3}}{2}\rightarrow x=\frac{-\sqrt{3}}{6}$$
(b)$$(2cos\Theta -1)(sin\Theta -1)=0$$
$$cos=\frac{1}{2} or sin\Theta =1$$
$$\Theta =\frac{\Pi }{3}$$ or $$\frac{\Pi }{2}$$