Cos\theta(\theta+\phi)=cos\theta is true only for \phi equal to 2\pi radians? I would like a proof or well explanation.

Question

`cos\theta(\theta+\phi)=cos\theta` is true only for
`\phi` equal to
`2\pi radians`? I would like a proof or well explanation.

Answer 1

Expand the left side using the angle sum identity for cosine:

\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phicos(θ+ϕ)=cosθcosϕ−sinθsinϕ

This reduces to \cos\thetacosθ if the sine product is 0 and \cos\phi=1cosϕ=1 . This happens when \phi=2n\piϕ=2nπ where nn is any integer; that is, any even multiple of \piπ .

So technically the claim is false; it's true for infinitely many values of \phiϕ . But it is certainly true for \phi=2\piϕ=2π rad.

Answer 2

Expand the left side using the angle sum identity for cosine:

`\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi`

This reduces to
`\cos\theta` if the sine product is 0 and
`\cos\phi=1`. This happens when
`\phi=2n\pi` where
`n` is any integer; that is, any even multiple of
`\pi`.

So technically the claim is false; it's true for infinitely many values of
`\phi`. But it is certainly true for
`\phi=2\pi` rad.