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Question

in triangle ABC there satisfies equation: $\cos A\cos B+\sin A\sin B\sin C=1$, determine possible values of C

What I have so far

I've noticed that the given equation looked similar to $\cos(A-B)$ which was $\cos A\cos B+\sin A\sin B$.

This I can extrapolate that $\cos(A-B)-\sin A\sin B+\sin A\sin B\sin C=1$

Thus $\cos(A-B)=1+\sin A\sin B+\sin A\sin B\sin C$

and when i factor out the common terms:

$\cos(A-B)=1+\sin A\sin B(1-\sin C)$

Right here I am not sure how to proceed and I got stuck

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marked as duplicate by lab bhattacharjee algebra-precalculus Jan 4 '17 at 18:10

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    $\begingroup$ Notice that $\cos(A-B) \leq 1$ and $\sin(A)\sin(B)(1-\sin(C)) \geq 0$ . $\endgroup$ – Blencer Jan 4 '17 at 14:21
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One can use the identity $$(\sin A-\sin B \sin C)^2+(\sin B\cos C)^2 + (\cos A-\cos B)^2$$ $$=2( 1 - ( \cos A\cos B + \sin A \sin B\sin C)) $$ to conclude $\cos C = 0$ and hence $\sin C = 1$. Thus $C=\frac {\pi }{2} $. Hope it helps.

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$\sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\leq 1$

$\cos A\cos B+\sin A\sin B\geq 1$

$\cos\left(A-B\right)\geq 1$

$\cos\left(A-B\right)=1$

$A-B=0$

$A=B$

$\sin C=\frac{1-\cos^2 A}{\sin^2 A}=1$

$C=90$

$A=B=45$

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