Prove that

Solution:

To Prove: $\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}=\frac{\sqrt{5}-1}{8}$

Taking LHS,

$=\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}$

We know that,

$\sin ^{2} A-\sin ^{2} B=\sin (A+B) \sin (A-B)$

$=\sin \left(24^{\circ}+6^{\circ}\right) \sin \left(24^{\circ}-6^{\circ}\right)$

$=\sin 30^{\circ} \sin 18^{\circ} \ldots(\mathrm{i})$

Now, we will find the value of sin 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.

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

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

$=\sin 30^{\circ} \sin 18^{\circ}$

$=\frac{1}{2} \times \frac{\sqrt{5}-1}{4}$

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

= RHS

∴ LHS = RHS

Hence Proved