prove identity: sin2a plus sin2b plus sin2c =4sina sinb sinc

Question

 a, b and c are angles of a triangle

Answer 1

An identity means that both sides are always equal regardless of the values for the variables.  We know that a+b+c = 180 since they are angles of a triangle. Since c = 180 - (a + b), then we also know that sin c = sin (180 - (a+b)).  It is easy to show that sin(180 - alpha) = sin(alpha) using the sum identity for sine.  Then sin c = sin (a+b). 

Also, cos(180 - alpha) = -cos(alpha) so cos c = = - cos(a+b)

 

Now sin(2a) = 2sin(a)cos(a) and sin(2b) = 2sin(b)cos(b)

Finally, sin(2c) = 2sin(c)cos(c) = -2sin(a+b)cos(a+b) = -2(sin a cos b + cos a sin b)(cos a cos b - sin a sin b)

= -2 [sin a cos a cos^2 b - cos a sin a sin^2 b + sin b cos b cos^2 a - sin^2 a sin b cos b]

= -2 [sin a cos a (cos^2 b - sin^2 b) + sin b cos b (cos^2 a - sin^2 a)]

= -2 [ sin a cos a (cos 2b) + sin b cos b (cos 2a)]

 

Putting this together on the left:

sin 2a + sin 2b + sin 2c

= 2 (sin a cos a + sin b cos b - sin(a+b)cos(a+b) )

= 2 (sin a cos a (1 - cos 2b) + sin b cos b (1 - cos 2a) )

 

 

On the right

4 sin a sin b sin c = 4 sin a sin b sin (a+b)

= 4 sin a sin b (sin a cos b + cos a sin b)

= 4 sin^2 a sin b cos b + 4 sin ^2 b sin a cos a

 

... well, I'm not sure how to equate these for now. I'll look at it more tomorrow.

 

... oh, I see, if you can show

4 sin^2 b = 2 (1 - cos 2b) and 4 sin ^2 a = 2(1 - cos 2a),

then you're done.

 

... and you can by using one of the the  cos (2 alpha) identities.

 

Good Luck.