We are given with,

…(i)

We need to find the value of

Take Left Hand Side (LHS) of equation (i),

Using the property of inverse trigonometry,

Putting and ,

Equate LHS to RHS.

Taking cosine on both sides,

Using property of inverse trigonometry,

cos(cos^-1 A) = A

Simplifying the equation,

Squaring on both sides,

Using algebraic identity,

(A – B)² = A² + B² – 2AB

Using trigonometric identity,

cos 2θ = cos² θ – sin² θ …(ii)

sin² θ + cos² θ = 1 ⇒ sin² θ = 1 – cos² θ …(iii)

Putting value of sin² θ from equation (iii) in equation (ii), we get

cos 2θ = cos² θ – (1 – cos² θ)

Or, cos 2θ = cos² θ – 1 + cos² θ

Or, cos 2θ = 2 cos² θ – 1

Or, 2 cos² θ = cos 2θ + 1

Replace θ by θ/2.

Substituting the value of in