Someone please help me

 see proof below

Step-by-step explanation:

Use the Double Angle Identity:  cos 2A = 2cos²A - 1

Use the Difference Identity:  cos (A - B) = cosA · cosB + sinA · sinB

Use the Unit Circle to evaluate:  cos (π/2) = 0   &     sin (π/2) = 1

Proof  LHS → RHS

LHS = RHS:   sin A = sin A  

Below.

Step-by-step explanation:

Left side = 2 cos^2 ( π/4 - A/2) - 1

= 2 ( cos π/4  cos A/2 + sin π/4  sin A/2)^2 - 1

Now sin π/4 and cos π/4 = 1 /√2 so:

= 2 ( 1/√2 cos A/2 + 1/√2 sin A/2)^2 - 1

= 2  * 1/2( cos^2 A/2 + sin^2 A/2 + 2 sin A/2 cos A/2) - 1

But cos^2 a/2 + sin^2 A/2 = 1 so we have:

2 * 1/2( 1 +  2sin A/2 cos A/2) - 1

= 1  + 2 sin A/2 cos A/2 - 1

=  2 sin A/2 cos A/2

Using the identity 2 sin A cos A = sin 2A

2 sin A/2 cos A/2 = sin A  = right side.

So left side = right side and the identity is proved.