Question

\[1+cos theta=2cos ^{2} theta/2\]

Answer 1

\[1+\cos\theta = 2\cos^2 (\frac{\theta}{2})\]

Are you to prove that?

Answer 2

yes

Answer 3

have you looked at the half-angle formula for cosine?

\[\cos^2 (u) = \frac{1+\cos(2u)}2\]

Answer 4

pls explain

Answer 5

\[\cos \theta = \cos ^{2}\frac{ \theta }{ 2 } - \sin ^{2}\frac{ \theta }{ 2 } \]

Answer 6

now put \[\sin ^{2}\frac{ \theta }{ 2 } = 1- \cos ^{2}\frac{ \theta }{ 2 } \]

Answer 7

let \(u = \frac{\theta}{2}\)

\[\cos^2(\frac{\theta}{2}) = \cos^2(u) = \frac{1+\cos(2u)}2\]\[2\cos^2(u) = 1+\cos(2u)\]\now undo the substitution
\[2\cos^2(\frac{\theta}{2}) = 1 + \cos(2*\frac{\theta}{2})\]\[2\cos^2(\frac{\theta}2) = 1+\cos(\theta)\]which is what we set out to demonstrate