#1-2sina*cosa÷2=sin^2(45^@-a)#?

1 Answer
Feb 15, 2018

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)