Question

A+B+C=3π/2,prove that cos2A+cos2B+cos2C=1-4 sinA sinB sinC

Answer 1

Step-by-step explanation:

We have,

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

Now,

$\sf{ cos(2A) + cos (2B) + cos(2C) }$

$\sf{ = 2cos \bigg( \dfrac{2A + 2B}{2} \bigg) cos \bigg( \dfrac{2A - 2B}{2} \bigg) + cos(2C) }$

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

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

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

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

$\sf{ =1 - 2sin( C ) \bigg\{ cos ( A - B) + sin \bigg( \dfrac{3\pi}{2} - (A + B)\bigg) \bigg \} }$

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

$\sf{ =1 - 2sin( C ) \cdot 2sin ( A ) sin ( B) }$

$\sf{ =1 - 4sin ( A ) sin ( B) sin( C ) }$