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Prove that if $x+y = z$ , then $\cos^2x + \cos^2y + \cos^2z - 2\cos x.\cos y.\cos z = \cos(x+y - z)$. My work: I tried to solve by converting the cos squared term into double angle form and transforming the product$- \cos x.\cos y$ into sum form. But not able to get the result.

Behrouz Maleki
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Rohit Sinha
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  • If $x+y = z$, then $\cos (x+y-z)$ is just $1$. Why not write that instead? –  Jul 08 '16 at 07:24

1 Answers1

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HINT:

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$\cos^2x+\cos^2y=\cdots=1+\cos(x+y)\cos(x-y)$$

Now if $\cos(x+y)=\cos z\iff x+y=2m\pi\pm z$ where $m$ is any integer

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lab bhattacharjee
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