prove the identity: 2cos^2 theta/2 = sin^2 theta/1-cos theta.
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calendarEducator since 2015
write85 answers
starTop subject is Math
Prove
`2cos^2(theta/2) = (sin^2theta)/(1-costheta)`
The half angle identity states:
`cos(theta/2) = sqrt((1+costheta)/2)`
This turns out left side into,
`2(1+costheta)/2 =(sin^2theta)/(1-costheta)`
`1+costheta =(sin^2theta)/(1-costheta)`
`(1-costheta)/(1-costheta) (1+costheta) =(sin^2theta)/(1-costheta)`
This simplifies into
`(1-cos^2theta)/(1-costheta) =(sin^2theta)/(1-costheta)`
Which turns into
`(sin^2theta)/(1-costheta) =(sin^2theta)/(1-costheta)`
It is proven!
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briefcaseCollege Professor
bookPh.D. from Oregon State University
calendarEducator since 2015
write3,394 answers
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Using the following identities:
`sin^2 x + cos^2 x =1`
and `cos^2 x = (1+cos 2x)/2`
Left hand side = 2 cos^2 (theta/2) = 2. (1+cos 2.theta/2)/2 = 1+cos theta
multiplying and dividing by (1-cos theta), we get.
LHS = (1+cos theta). (1-cos theta)/ (1-cos theta) = (1-cos^2 theta)/(1-cos theta)
= (sin^2 theta + cos^2 theta - cos theta)/(1cos theta) = sin^2 theta/ (1-cos theta). = RHS
Hence proved.
using half angle identities
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