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In $\Delta ABC$,if $$\cos{C}\cdot(\sin{A}+\sin{B})=\sin{C}\cdot\cos{(A-B)}$$ Find $\cos{A}+\cos{B}$

Thus $$\sin{A}+\sin{B}=\tan{C}\cdot\cos{(A-B)}=\dfrac{\sin{(A+B)}}{\cos{(A+B)}}\cos{(A-B)}$$ $$(\sin{A}+\sin{B})\cos{(A+B)}=\sin{(A+B)}\cos{(A-B)}$$ follow I can't figure it out.

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    I think it should be $$\tan(C)=\frac{\sin(\pi-(A+B))}{\cos(\pi-(A+B))}=\frac{\sin(A+B)}{-\cos(A+B)}$$ – awllower Aug 10 '15 at 11:14

1 Answers1

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Let me try. $\cos C (\sin A +\sin B)=\sin C \cos(A-B)$

$\implies2\left(1-2\sin^2\dfrac C2\right)\sin\dfrac{A+B}2\cos\dfrac{A-B}2 = 2\sin \dfrac C2\cos\dfrac C2\left(2\cos^2\dfrac{A-B}2-1\right)\ \ \ \ (1)$

Now as $\sin\dfrac{A+B}2=\sin\left(\dfrac\pi2-\dfrac C2\right)=\cos\dfrac C2$

and $\cos\dfrac C2=0\implies\dfrac C2=(2n+1)\dfrac\pi2\iff C=(2n+1)\pi$ where $n$ is any integer

But $0<C<\pi\implies\sin\dfrac{A+B}2=\cos\dfrac C2\ne0$

$(1)\implies$

$$\left(1-2\cos^2\dfrac{A+B}2\right)\cos\dfrac{A-B}2 = \cos\dfrac{A+B}2\left(2\cos^2\dfrac{A-B}2-1\right)$$

$$\implies\left(\cos\dfrac{A-B}2+\cos\dfrac{A+B}2\right)\left(1-2\cos\dfrac{A-B}2\cos\dfrac{A+B}2\right)=0$$

Now, $\cos\dfrac{A-B}2+\cos\dfrac{A+B}2=2\cos\dfrac A2\cos\dfrac B2$ which can not be zero due to the reason mentioned above.

$$\implies1-2\cos\dfrac{A-B}2\cos\dfrac{A+B}2=0$$

But $2\cos\dfrac{A-B}2\cos\dfrac{A+B}2=\cos A+\cos B$

lab bhattacharjee
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GAVD
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  • Thanks for this great answer. It is intriguing to see such a use of these identities. P.S. Now I see how to deduce the last equation, so just thanks. :-) – awllower Aug 10 '15 at 11:44
  • Though I think it would be better to add that $\cos(\frac{A-B}{2}+\sin(\frac{C}{2}))=2\cos(\frac{A}{2})\cos(\frac{B}{2})\not=0.$ – awllower Aug 10 '15 at 11:51
  • @awllower: Ah yes, because $\frac{C}{2} >0$ and $\frac{A-B}{2} < \frac{\pi}{2}$, so $\cos(\frac{A-B}{2}) + \sin(\frac{C}{2}) \neq 0$ – GAVD Aug 11 '15 at 02:02
  • @GAVD, Please verify the updated answer. – lab bhattacharjee Aug 11 '15 at 16:52
  • @labbhattacharjee Uhm thanks you for the details. I just gave an idea and it should be improved as your edits. – GAVD Aug 11 '15 at 16:58