Question

How do you show that `4cos^2x- 2 = 2sin(2x+ pi/2)`?

Answer 1

`=>2(2cos^2x - 1) = 2sinx[2(x+ pi/4)]`

`=>2cos^2x- 1 = sin[2(x + pi/4)]`

Make `x + pi/4 = t`.

`=>2cos^2x - 1 = sin(2t)`

By the formula `sin2x = 2sinxcosx`:

`=>2cos^2x- 1 = 2sintcost`

`=>2cos^2x - 1 = 2sin(x + pi/4)cos(x+ pi/4)`

Expand using the sum formulae `sin(A + B) = sinAcosB + cosAsinB` and `cos(A + B) = cosAcosB - sinAsinB`.

`=>2cos^2x -1 = 2(sinxcos(pi/4) + cosxsin(pi/4))(cosxcos(pi/4) - sinxsin(pi/4))`

`=>2cos^2x - 1 = 2(sinx(1/sqrt(2)) + cosx(1/sqrt(2))(cosx(1/sqrt(2)) - sinx(1/sqrt(2)))`

`=>2cos^2x - 1 = 2((sinx + cosx)/sqrt(2))((cosx - sinx)/sqrt(2))`

`=>2cos^2x- 1 = 2((cos^2x - sin^2x))/2`

`=>2cos^2x- 1 = cos^2x- sin^2x`

Use `sin^2theta + cos^2theta = 1`:

`=>2cos^2x - 1 = cos^2x + cos^2x - 1`

`=>2cos^2x - 1 = 2cos^2x - 1`

`LHS = RHS`

Identity proved!!

Hopefully this helps!