Question

The value of the expression ${\left(\frac{\sin 3\theta }{\sin \theta }\right)}^2-{\left(\frac{\cos 3\theta }{\cos \theta }\right)}^2$,when $\theta =(7.5{)}^o$, is

• A. $4(\sqrt{3}+1)$
• B. $(\sqrt{3}-1)$
• C. $2(\sqrt{6}+\sqrt{2})$
• D. $\frac{8}{\sqrt{6}-\sqrt{2}}$

Answer 1

Answer: (C)

$2(\sqrt{6}+\sqrt{2})$

$\frac{\sin {}^{3}3\theta }{\sin {}^2\theta }-\frac{\cos {}^{2}3\theta }{\cos {}^2\theta }(\theta ={7.5}^{\circ })$

We have,

$\sin 3\theta =3\sin \theta -4\sin {}^3\theta \cos 3\theta =4\cos {}^3\theta -3\cos \theta \frac{\sin 3\theta }{\sin \theta }=3-4\sin {}^2\theta \frac{\cos 3\theta }{\cos \theta }=4\cos {}^2\theta -3\therefore \frac{\sin {}^{2}3\theta }{\sin {}^2\theta }-\frac{\cos {}^{2}3\theta }{\cos {}^2\theta }={(3-4\sin {}^2\theta )}^2-{(4\cos {}^2\theta -3)}^2=(9+16\sin {}^4\theta -24\sin {}^2\theta )-(9+16\cos {}^4\theta -24\cos {}^2\theta )=24(\cos {}^2\theta -\sin {}^2\theta )-16(\cos {}^4\theta -\sin {}^4\theta )=24(\cos {}^2\theta -\sin {}^2\theta )-16(\cos {}^2\theta +\sin {}^2\theta )(\cos {}^2\theta -\sin {}^2\theta )=24(\cos {}^2\theta -\sin {}^2\theta )-16(\cos {}^2\theta -\sin {}^2\theta )=8(\cos {}^2\theta -\sin {}^2\theta )=8\cos 2\theta =8\cos {15}^{\circ }=8\cos ({45}^{\circ }-{35}^{\circ })=8[\cos {45}^{\circ }.\cos {30}^{\circ }+\sin {45}^{\circ }.\sin {30}^{\circ }]=8[\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}]=\frac{4(\sqrt{3}+1)}{\sqrt{2}}=2\sqrt{2}(\sqrt{3}+1)=2(\sqrt{6}+\sqrt{2})$