Question

`1-2sina*cosa÷2=sin^2(45^@-a)`?

Answer 1

See below for a possible answer.

Explanation:

`sin^2(45-a)=sin(45-a)sin(45-a)`

Using the angle subtraction formula for sine:

`sin(45-a)sin(45-a)=(sin45cosa-sinacos45)(sin45cosa-sinacos45)`

`sin45=cos45=sqrt2/2` So:

`(sqrt2/2cosa-sqrt2/2sina)(sqrt2/2cosa-sqrt2/2sina)`

Factor out the `sqrt2/2` in both expressions:

`sqrt2/2(cosa-sina)sqrt2/2(cosa-sina)`

Multiply the square roots and the trig expressions:

`2/4(cos^2a-2sinacosa+sin^2a)`

Simplify fraction and rearrange in parentheses:

`1/2(sin^2a+cos^2a-2sinacosa)`

`sin^2a+cos^2a=1` therefore:

`sin^2(45-a)=(1-2sinacosa)/2`

QED

(Don't know how to get the degrees mark. I'll edit it if I figure it out, but all of the 45s above are 45degrees)