1-2sina*cosa÷2=sin^2(45^@-a)12sinacosa÷2=sin2(45a)?

1 Answer
Feb 15, 2018

See below for a possible answer.

Explanation:

sin^2(45-a)=sin(45-a)sin(45-a)sin2(45a)=sin(45a)sin(45a)

Using the angle subtraction formula for sine:

sin(45-a)sin(45-a)=(sin45cosa-sinacos45)(sin45cosa-sinacos45)sin(45a)sin(45a)=(sin45cosasinacos45)(sin45cosasinacos45)

sin45=cos45=sqrt2/2sin45=cos45=22 So:

(sqrt2/2cosa-sqrt2/2sina)(sqrt2/2cosa-sqrt2/2sina)(22cosa22sina)(22cosa22sina)

Factor out the sqrt2/222 in both expressions:

sqrt2/2(cosa-sina)sqrt2/2(cosa-sina)22(cosasina)22(cosasina)

Multiply the square roots and the trig expressions:

2/4(cos^2a-2sinacosa+sin^2a)24(cos2a2sinacosa+sin2a)

Simplify fraction and rearrange in parentheses:

1/2(sin^2a+cos^2a-2sinacosa)12(sin2a+cos2a2sinacosa)

sin^2a+cos^2a=1sin2a+cos2a=1 therefore:

sin^2(45-a)=(1-2sinacosa)/2sin2(45a)=12sinacosa2

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)