Given $A + B + C = 180$, prove that $$\sin^2(A) + \sin^2(B) - \sin^2(C) = 2\sin(A)\sin(B)\cos(C).$$ I tried all identities I know but I have no idea how to proceed.
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The given formula looks a lot like the law of cosines - and we can use this quite elegantly.
Note that we can interpret $A$, $B$, and $C$ as the angles in a triangle with side lengths $a$, $b$ and $c$. We can choose $a$ as anything we want, so let $a=\sin(A)$, and by the law of sines:
$$\frac{\sin (A)}{a}=\frac{\sin (B)}{b}=\frac{\sin (C)}{c}$$
we have a triangle with angles $A$, $B$, and $C$ such that $a=\sin (A)$, $b=\sin (B)$, and $c=\sin (C)$
Now we can apply the law of cosines:
$$ c^2=a^2+b^2-2ab\cos(C) \implies \sin^2 (C)=\sin^2(A)+\sin^2(B)-2\sin(A)\sin(B)\cos(C) $$
Rearranging: voila, we get the desired equality!
$$\sin^2(A)+\sin^2(B)-\sin^2 (C)=2\sin(A)\sin(B)\cos(C)$$
Peter Woolfitt
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Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
and $$\sin A=\sin[\pi-(B+C)]=\sin(B+C)$$
$$\sin^2A+\sin^2B-\sin^2C$$ $$=\sin^2 A+\sin(B-C)\sin(B+C)$$
$$=\sin A[\sin A+\sin(B-C)]$$
$$=\sin A[\sin(B+C)+\sin(B-C)]$$
$$=\sin A[2\sin B\cos C]$$ $$=2\sin A \sin B \cos C$$
lab bhattacharjee
- 274,582
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Can you please elaborate a little bit? – Kanishk Jul 25 '14 at 06:49
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@Kanishk, Please find the edited answer – lab bhattacharjee Jul 25 '14 at 08:16