Question

How do you prove `sin^(2)(x+π/4)-sin^(2)(x-π/4)=2sinx*cosx`?

Answer 1

See proof.

Explanation:

We can prove the identity if we can start with the left-hand side and end up at the right side.

Along the way, let's use the following trigonometric identities:

[1] `" " sin(x+y) = sinx cos y + cos x sin y`

[2] `" " sin(x-y) = sinx cos y - cos x sin y`

and the binomial formula

[3] `" "(a + b)^2 = a^2 + 2ab + b^2`

[4] `" "(a - b)^2 = a^2 - 2ab + b^2`

So, let's start:

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

`stackrel("[1],[2]")(" "=) [sin(x) cos(pi/4) + cos(x) sin(pi/4)]^2`

`- [sin(x) cos(pi/4) - cos(x) sin(pi/4)]^2`

`" " stackrel("[3],[4]")(" "=) [sin^2x cos^2(pi/4) + 2 sinx cosx sin(pi/4)cos(pi/4) + cos^2x sin^2(pi/4)]`

`- [sin^2 x cos^2(pi/4) - 2sinx cosx sin(pi/4)cos(pi/4) + cos^2 x sin^2(pi/4)]`

`" " = color(blue)( cancel(sin^2x cos^2(pi/4))) + 2 sinx cosx sin(pi/4)cos(pi/4)+ color(red)(cancel(cos^2 x sin^2(pi/4)))`

`color(blue)(cancel(- sin^2 x cos^2(pi/4))) + 2sinx cosx sin(pi/4)cos(pi/4) color(red)(cancel(- cos^2 x sin^2(pi/4)))`

`= 4 sin x cos x sin(pi/4)cos(pi/4)`

... remember that `sin(pi/4) = cos(pi/4) = sqrt(2)/2`...

`= 4sin x cos x * sqrt(2)/2*sqrt(2)/2`

`= cancel(4)sin x cos x * 2/cancel(4)`

`= 2sin x cos x`

As we have safely arrived at the right-hand side, we haven proven the identity.

q.e.d.

I wasn't able to format the solution in a better way or find a way to have the whole term on one line during some steps. I hope that this is readable enough nevertheless.