Question

How do you verify `(sin B + cos B)(sin B - cos B)= 2 sin^2B - 1`?

Answer 1

Have a look:

Explanation:

We can multiply the brackets and get:
`sin^2Bcancel(-sinBcosB)cancel(+sinBcosB)-cos^2B=2sin^2B-1`
`sin^2B-cos^2B=2sin^2B-1`
But: `sin^2B+cos^2B=1`
So:
`sin^2B-cos^2B=2sin^2B-sin^2B-cos^2B`
And:
`sin^2B-2sin^2B+sin^2B=cos^2B-cos^2B`
`0=0`

Answer 2

Verify `(sin B - cos B)(sin B + cos B) = 2sin^2 B - 1`

Explanation:

`sin^2 B - cos^2 B = 2sin^2 B - 1`

`- (cos^2B - sin^2 B) = - (1 - 2sin^2 B)`

`- cos (2B) = - cos (2B)`

Reminder: `cos 2a = 1 - 2sin^2a = cos^2 a - sin^2 a`