Answer :

To Prove:

Taking LHS,

= sin^{2}24° - sin^{2}6°

We know that,

sin^{2}A – sin^{2}B = sin(A + B) sin(A – B)

= sin(24°+ 6°) sin(24° - 6°)

= sin 30° sin 18° …(i)

Now, we will find the value of sin 18°

Let x = 18°

so, 5x = 90°

Now, we can write

2x + 3x = 90°

so 2x = 90° - 3x

Now taking sin both the sides, we get

sin2x = sin(90° - 3x)

sin2x = cos3x [as we know, sin(90°- 3x) = Cos3x ]

We know that,

sin2x = 2sinxcosx

Cos3x = 4cos^{3}x - 3cosx

2sinxcosx = 4cos^{3}x - 3cosx

⇒ 2sinxcosx - 4cos^{3}x + 3cosx = 0

⇒ cosx (2sinx - 4cos^{2}x + 3) = 0

Now dividing both side by cosx we get,

2sinx - 4cos^{2}x + 3 = 0

We know that,

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

or cos^{2}x = 1 – sin^{2}x

⇒ 2sinx – 4(1 – sin^{2}x) + 3 = 0

⇒ 2sinx – 4 + 4sin^{2}x + 3 = 0

⇒ 2sinx + 4sin^{2}x – 1 = 0

We can write it as,

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

Now applying formula

Here, ax^{2} + bx + c = 0

So,

now applying it in the equation

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

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

= sin 30° sin 18°

= RHS

∴ LHS = RHS

Hence Proved

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Prove that:

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