1-cos(theta)=sin^2(theta)/(1+cos(theta))1−cos(θ)=sin2(θ)1+cos(θ)
Recall that sin^2(theta)+cos^2(theta)=1sin2(θ)+cos2(θ)=1. Solving for sin^2(theta)sin2(θ) yields:
sin^2(theta)=1-cos^2(theta)sin2(θ)=1−cos2(θ)
Rewrite the right side using the above identity:
1-cos(theta)=(1-cos^2(theta))/(1+cos(theta))1−cos(θ)=1−cos2(θ)1+cos(θ)
Recall the difference of squares, which states that (x^2-a^2)=(x+a)(x-a)(x2−a2)=(x+a)(x−a)
Similarly, (a^2-x^2)=(a+x)(a-x)(a2−x2)=(a+x)(a−x)
For (1-cos^2(theta)),(1−cos2(θ)), we can use difference of squares where a^2=1, x^2=cos^2(theta), a=sqrt(1)=1, x=sqrt(cos^2(theta))=cos(theta):a2=1,x2=cos2(θ),a=√1=1,x=√cos2(θ)=cos(θ):
(1-cos^2(theta))=(1+cos(theta))(1-cos(theta))(1−cos2(θ))=(1+cos(θ))(1−cos(θ))
Rewrite the right side again after applying this. (1+cos(theta))(1+cos(θ)) cancels:
1-cos(theta)=(cancel(1+cos(theta))(1-cos(theta)))/cancel(1+cos(theta))
1-cos(theta)=1-cos(theta)