If alpha+beta=gamma,then what is the value of | vedclass

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(C) Given,$\alpha+\beta=\gamma$. We need to evaluate $\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma$. Using the identity $2 \cos^2 	heta = 1+\cos 2 	heta

Vedclass · Accepted Answer

$1+2 \cos \alpha \cos \beta \cos \gamma$

Vedclass · Answer

(C) Given,$\alpha+\beta=\gamma$. We need to evaluate $\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma$. Using the identity $2 \cos^2 	heta = 1+\cos 2 	heta$,we have: $\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma = \frac{1}{2} [2 \cos^2 \alpha + 2 \cos^2 \beta + 2 \cos^2 \gamma]$ $= \frac{1}{2} [1+\cos 2 \alpha + 1+\cos 2 \beta + 2 \cos^2 \gamma]$ $= \frac{1}{2} [2 + 2 \cos(\alpha+\beta) \cos(\alpha-\beta) + 2 \cos^2 \gamma]$ Since $\alpha+\beta=\gamma$,substitute $\gamma$ for $\alpha+\beta$: $= \frac{1}{2} [2 + 2 \cos \gamma \cos(\alpha-\beta) + 2 \cos^2 \gamma]$ $= 1 + \cos \gamma \cos(\alpha-\beta) + \cos^2 \gamma$ $= 1 + \cos \gamma [\cos(\alpha-\beta) + \cos \gamma]$ $= 1 + \cos \gamma [\cos(\alpha-\beta) + \cos(\alpha+\beta)]$ Using $\cos(A-B) + \cos(A+B) = 2 \cos A \cos B$: $= 1 + \cos \gamma [2 \cos \alpha \cos \beta]$ $= 1 + 2 \cos \alpha \cos \beta \cos \gamma$.

If $\alpha+\beta=\gamma$,then what is the value of $\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma$ equal to?

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