Use trig identites to transform the left side of equation into the right. (1+cos theta)(1-cos theta)= sin^2theta
1+cos theta =2cos^2 theta/2
multiply on the left, and get \[1-\cos^2(\theta)\] then it should be easy enough
1-cos theta=2sin^2 theta/2
is it clear that \[1-\cos^2(\theta)=\sin^2(\theta)\]?
ooohh. yea. I see.
kk good
thank you(:
yw
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use identity ---> a^2 - b^2 for lhs then prove using cos2 + sin2 = 1
(1+cosθ)(1−cosθ)=(1−cosθ+cosθ−cos2θ) = sin2theta
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