We need to find the value of

Let

Now, take sine on both sides,

Using the property of inverse trigonometry,

sin(sin^-1 A) = A

Let us find the value of cos x.

We know by trigonometric identity, that

sin² x + cos² x = 1

⇒ cos² x = 1 – sin² x

Put the value of sin x,

We have,

…(i)

Using the trigonometric identity,

cos 2x = cos² x – sin² x

⇒ cos 2x = (1 – sin²x) – sin² x [∵, sin² x + cos² x = 1]

⇒ cos 2x = 1 – sin² x – sin² x

⇒ cos 2x = 1 – 2 sin² x

Or,

2 sin² x = 1 – cos 2x

Replacing x by x/4,

Substituting the value of in equation (i),

…(ii)

Using the trigonometric identity,

cos 2x = cos² x – sin² x

⇒ cos 2x = cos² x – (1 – cos² x) [∵, sin² x + cos² x = 1]

⇒ cos 2x = cos² x – 1 + cos²x

⇒ cos 2x = 2 cos² x – 1

Or,

2 cos² x = 1 + cos 2x

Replacing x by x/2,

Substituting the value of in equation (ii),

Put the value of cos x as found above, cos x = 1/8.