sin^2(45-a)=sin(45-a)sin(45-a)sin2(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)sin(45−a)sin(45−a)=(sin45cosa−sinacos45)(sin45cosa−sinacos45)
sin45=cos45=sqrt2/2sin45=cos45=√22 So:
(sqrt2/2cosa-sqrt2/2sina)(sqrt2/2cosa-sqrt2/2sina)(√22cosa−√22sina)(√22cosa−√22sina)
Factor out the sqrt2/2√22 in both expressions:
sqrt2/2(cosa-sina)sqrt2/2(cosa-sina)√22(cosa−sina)√22(cosa−sina)
Multiply the square roots and the trig expressions:
2/4(cos^2a-2sinacosa+sin^2a)24(cos2a−2sinacosa+sin2a)
Simplify fraction and rearrange in parentheses:
1/2(sin^2a+cos^2a-2sinacosa)12(sin2a+cos2a−2sinacosa)
sin^2a+cos^2a=1sin2a+cos2a=1 therefore:
sin^2(45-a)=(1-2sinacosa)/2sin2(45−a)=1−2sinacosa2
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)