Question

if a+b+c=3π/2,then prove that 1)cos^2a+cos^2b-cos^2c=-2cosAcosBsinC ​

Answer 1

Step-by-step explanation:

We have,

$A + B + C = \frac{3\pi}{2}$

Now,

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

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

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

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

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

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

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

$\rm \: = \frac{1}{2} + \frac{1}{2} \bigg \{2\cos \bigg( \frac{3\pi}{2} - C \bigg) \cos ( A - B) + 2\sin^{2} (C) \bigg \} - \frac{1}{2}$

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

$\rm \: = \frac{1}{2} \bigg \{ 2\sin ( C ) \{ \sin(C) - \cos ( A - B) \} \bigg \}$

$\rm \: = \sin ( C ) \bigg\{ \sin \bigg( \frac{3\pi}{2} - ( A + B)\bigg) - \cos ( A - B) \bigg\}$

$\rm \: = \sin ( C ) \bigg\{ - \cos( A + B) - \cos ( A - B) \bigg\}$

$\rm \: = - \sin ( C ) \bigg\{ \cos( A + B) + \cos ( A - B) \bigg\}$

$\rm \: = - \sin ( C ) \bigg\{ 2\cos( A )\cos ( B) \bigg\}$

$\rm \: = - 2\cos( A )\cos ( B) \sin ( C )$