Question

If A, B, C are angles in a triangle, then prove that cos 2A - cos 2B + cos 2C = 1 - 4 sin A cos B sin C.

Answer 1

Answer:

Step-by-step explanation:

Formula used:

$cosC-cosD=-2\:sin(\frac{C+D}{2}).sin(\frac{C-D}{2})\\sin2A=1-2{sin}^2C$

$cos2A-cos2B+cos2C$

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

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

$=-2\:sin(A-B).sin(180-C)+1-2{sin}^2C$

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

=1-2sinC[sin(A-B)+sinC]

=1-2sinC[sin(A-B)+sin(180-(A+B))]

=1-2sinC[sin(A-B)+sin(A+B)]

=1-2sinC[2 sinA cosB]

=1- 4sinA cosB sinC

Answer 2

Answer:

Formula used:

=1-2sinC[sin(A-B)+sinC]

=1-2sinC[sin(A-B)+sin(180-(A+B))]

=1-2sinC[sin(A-B)+sin(A+B)]

=1-2sinC[2 sinA cosB]

=1- 4sinA cosB sinC

Step-by-step explanation:

HOPE IT IS HELPFUL