Question

If $A+B+C={180}^o$ then prove the following:
(i)${\sin }^{2}A+{\sin }^{2}B-{\sin }^{2}C=2\sin A\sin B\cos C$
(ii) ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2(1+\cos A\cos B\cos C)$

Answer 1

We have

$A+B+C={180}^o$ …… (1)

Prove that:-

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

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

Proof:-

Part (1).. L.H.S.

${\sin }^{2}A+{\sin }^{2}B-{\sin }^{2}C$

$=\left(\frac{1-\cos 2A}{2}\right)+\left(\frac{1-\cos 2A}{2}\right)-(1-{\cos }^{2}C)\therefore \cos 2\theta =1-2si{n}^2\theta$

$=1-\frac{1}{2}(\cos 2A+\cos 2B)+{\cos }^{2}C-1$

$=-\frac{1}{2}(\cos 2A+\cos 2B)+{\cos }^{2}C\therefore \cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}$

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

$=-\left[\cos (A+B)\cos (A-B)\right]+{\cos }^{2}C$

$=-\left[\cos ({180}^o-C)\cos (A-B)\right]+{\cos }^{2}C\therefore \cos ({180}^o-\theta )=-\cos \theta$

$=-[-\cos C\cos (A-B)]+{\cos }^{2}C$

$=\cos C\cos (A-B)+{\cos }^{2}C$

$=\cos C[\cos (A-B)+\cos C]$

$=\cos C[\cos (A-B)+\cos ({180}^o-(A+B))]\therefore \cos ({180}^o-\theta )=-\cos \theta$

$=\cos C[\cos (A-B)-\cos (A+B)]$

$=\cos C\left(2\sin A\sin B\right)\therefore 2\sin A\sin B=\cos (A-B)-\cos (A+B)$

$=2\sin A\sin B\cos C$

R.H.S.

Hence proved.

Part (2).

L.H.S.

${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$

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

$=1-({\cos }^{2}A-{\sin }^{2}B)+(1-{\cos }^{2}C)$

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

$=2-\left(\cos [\pi -C]\cos (A-B)\right)-{\cos }^{2}C$

$=2-(-\cos C\cos (A-B))-{\cos }^{2}C$

$=2+\cos C\cos (A-B)-{\cos }^{2}C$

$=2+\cos C(\cos (A-B)-\cos C)$

$=2+\cos C(\cos (A-B)-\cos C)$

$=2+\cos C(\cos (A-B)-\cos [{180}^o-(A+B)])$

$=2+\cos C(\cos (A-B)-\cos (A+B))$

$=2+\cos C\left(2\cos A\cos B\right)$

$=2+2\cos A\cos B\cos C$

$=2(1+\cos A\cos B\cos C)$

R.H.S.

Hence proved.