Prove that

To Prove: $\sin ^{2} 72^{\circ}-\cos ^{2} 30^{\circ}=\frac{\sqrt{5}-1}{8}$

Taking LHS,

$=\sin ^{2} 72^{\circ}-\cos ^{2} 30^{\circ}$

$=\sin ^{2}\left(90^{\circ}-18^{\circ}\right)-\cos ^{2} 30^{\circ}$

$=\cos ^{2} 18^{\circ}-\cos ^{2} 30^{\circ} \ldots$ (i)

Here, we don’t know the value of cos 18°. So, we have to find the value of cos 18°

Let $x=18^{\circ}$

so, $5 x=90^{\circ}$

Now, we can write

$2 x+3 x=90^{\circ}$

so $2 x=90^{\circ}-3 x$

Now taking sin both the sides, we get

$\sin 2 x=\sin \left(90^{\circ}-3 x\right)$

$\sin 2 x=\cos 3 x\left[\right.$ as we know, $\left.\sin \left(90^{\circ}-3 x\right)=\cos 3 x\right]$

We know that,

$\sin 2 x=2 \sin x \cos x$

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

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

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

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

Now dividing both side by cosx we get

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

We know that,

$\cos ^{2} x+\sin ^{2} x=1$

or $\cos ^{2} x=1-\sin ^{2} x$

$\Rightarrow 2 \sin x-4\left(1-\sin ^{2} x\right)+3=0$

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

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

We can write it as,

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

Now applying formula

Here, $a x^{2}+b x+c=0$

So, $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

now applying it in the equation

$\sin x=\frac{-2 \pm \sqrt{2^{2}-4(4)(-1)}}{2}$

$\sin x=\frac{-2 \pm \sqrt{4+16}}{8}$

$\sin x=\frac{-2 \pm \sqrt{20}}{8}$

$\sin x=\frac{(-2 \pm 2 \cdot \sqrt{5})}{8}$

$\sin x=\frac{2(-1 \pm \sqrt{5})}{8}$

$\sin x=\frac{-1 \pm \sqrt{5}}{4}$

$\sin 18^{\circ}=\frac{-1 \pm \sqrt{5}}{4}$

Now sin 18° is positive, as 18° lies in first quadrant.

$\therefore \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$

Now, we know that

$\cos ^{2} x+\sin ^{2} x=1$

or $\cos x=\sqrt{1}-\sin ^{2} x$

$\therefore \cos 18^{\circ}=\sqrt{1}-\sin ^{2} 18^{\circ}$

$\Rightarrow \cos 18^{\circ}=\sqrt{1-\left(\frac{\sqrt{5}-1}{4}\right)^{2}}$

$\Rightarrow \cos 18^{\circ}=\sqrt{\frac{16-(5+1-2 \sqrt{5})}{16}}$

$\Rightarrow \cos 18^{\circ}=\sqrt{\frac{16-6+2 \sqrt{5}}{16}}$

$\Rightarrow \cos 18^{\circ}=\frac{1}{4} \sqrt{10+2 \sqrt{5}}$

Putting the value in eq. (i), we get

$=\cos ^{2} 18^{\circ}-\cos ^{2} 30^{\circ}$

$=\left(\frac{1}{4} \sqrt{10+2 \sqrt{5}}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\left[\because \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]$

$=\frac{1}{16}(10+2 \sqrt{5})-\frac{3}{4}$

$=\frac{10+2 \sqrt{5}-12}{16}$

$=\frac{2 \sqrt{5}-2}{16}$

$=\frac{2(\sqrt{5}-1)}{16}$

$=\frac{\sqrt{5}-1}{8}$

= RHS

∴ LHS = RHS

Hence Proved