NIMCET 2017 — Mathematics PYQ

Concept:

$\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

$\sin 2\theta =2\sin \theta \cos \theta$

$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta =1-2{\sin }^2\theta$

Calculation:

Let us consider $B=18^{\circ }$

So, $5B=90^{\circ }$

$\Rightarrow 2B+3B=90^{\circ }$

$\Rightarrow 2B=90^{\circ }-3B$

By taking sine on both sides, we get

$\sin 2B=\sin (90^{\circ }-3B)=\cos 3B$

$\Rightarrow 2\sin B\cos B=4{\cos }^{3}B-3\cos B$

$\Rightarrow 2\sin B\cos B-4{\cos }^{3}B+3\cos B=0$

$\Rightarrow \cos B(2\sin B-4{\cos }^{2}B+3)=0$

Dividing both sides by $\cos B$ (as $\cos 18^{\circ }\ne 0$), we get

$2\sin B-4(1-{\sin }^{2}B)+3=0$

$\Rightarrow 4{\sin }^{2}B+2\sin B-1=0$

This is a quadratic equation in sine.

So, $\sin B=\frac{-2\pm \sqrt{4+16}}{2\times 4}=\frac{-1\pm \sqrt{5}}{4}$

Now, since $18^{\circ }$ is in the first quadrant, $\sin 18^{\circ }$ is positive.

Therefore, $\sin B=\frac{-1+\sqrt{5}}{4}$

Now, $\cos 36^{\circ }=\cos (2\times 18^{\circ })$

$\Rightarrow \cos 36^{\circ }=1-2{\sin }^{2}18^{\circ }$

$\Rightarrow \cos 36^{\circ }=1-2{\left(\frac{\sqrt{5}-1}{4}\right)}^2=\left(\frac{16-2\times (5+1-2\sqrt{5})}{16}\right)=\frac{4+4\sqrt{5}}{16}$

$\Rightarrow \cos 36^{\circ }=\frac{\sqrt{5}+1}{4}$

As we know, ${\sin }^2\theta +{\cos }^2\theta =1$

$\Rightarrow \sin \theta =\sqrt{1-{\cos }^2\theta }$

Now, $\sin 36^{\circ }=\sqrt{1-{\left(\frac{\sqrt{5}+1}{4}\right)}^2}$

$\Rightarrow \sin 36^{\circ }=\frac{\sqrt{10-2\sqrt{5}}}{4}$