Trigonometric Identity?

Question

Given that A + B + C = 180°, prove the identity.

sin 2A + sin 2B + sin 2C = 4(sin A)(sin B)(sin C)

Answer 1

Note: A + B + C = 180 degrees.

sin[2A] + sin[2B] + sin[2(180-A-B)] ... left side
4sin[A] sin[B] sin[C] = 4sin[A] sin[B] sin[(180-A-B)] .. right side

§ start with left side:

sin[2A] + sin[2B] + sin[360 - 2(A+B)]
= sin[2A] + sin[2B] + sin[ -(2A+2B)]
= sin[2A] + sin[2B] - (sin[2A] cos[2B] + cos[2A] sin[2B])
= sin[2A] {1 - cos[2B]} + sin[2B] {1 - cos[2A]}
= sin[2A] {1 - 1 + 2sin²[B]} + sin[2B] {1 - 1 + 2sin²[A]}
= 2sin[2A] sin²[B] + 2sin[2B] sin²[A]
= 4sin[A]cos[A] sin²[B] + 4sin[B] cos[B] sin²[A]
= 4 sin[A]sin[B] {cos[A]sin[B] + sin[A] cos[B]}
= 4 sin[A] sin[B] sin[A+B]
= 4 sin[A] sin[B] sin[180 - (A+B)]
= 4 sin[A] sin[B] sin[C]

Answer 2

C = 180 - (A + B)
2C = 360 - (2A + 2B)
sin C = sin (A + B); sin 2C = - sin (2A + 2B)
sin 2A + sin 2B + sin 2C = sin 2A + sin 2B - sin (2A + 2B)
= sin 2A + sin 2B - sin 2A * cos 2B - cos 2A * sin 2B = sin 2A * (1 - cos 2B) + sin 2B * ( 1 - cos 2A) = sin 2A * 2 * (sin B)^2 + sin 2B * 2 * (sin A)^2 = 2 * sin A * cos A * 2 * (sin B)^2 + 2 * sin B * cos B *2 * (sin A)^2 = 4 * sin A * sin B * ( sin B * cos A + cos A * sin B) = 4 * sin A * sin B * sin (A + B) = 4 * sin A * sin B * sin C

Answer 3

Hmm..To start off, lets combine sin2A and sin2B. using the identity sinA+ sinB= 2sin((A+B)/2)cos((A-B)/2), we have:
sin 2A + sin 2B = 2sin (A+B) cos(A-B),
since A+B = 180-C, and sin(180-C) = sinC, we have:
sin 2A + sin 2B = 2sinCcos(A-B). now, we add sin2C.
As we know, sin2C = 2sinC cosC,
sin2A+sin2B+sin2C = 2sinC cos(A-B) + 2sinC cosC.
factoring 2sinC, we get:
=2sinC( cos(A-B) + cos C).
using the identity cosA + cosB = 2cos((A+B)/2)cos((A-B)/2),
we have cos(A-B)+cos C = 2cos ((A-B+C)/2) cos((A-B-C)/2)
so, we now have:
=2sinC X 2cos ((A-B+C)/2) cos((A-B-C)/2).
since A+C = 180-B,
cos ((A-B+C)/2) becomes cos ((180-2B)/2) = cos(90-B)
which is simply sinB
also, since B+C= 180-A,
cos((A-B-C)/2) becomes cos((A-180+A)/2) = cos (A-90) = cos(90-A) since cosine is an even function which is sinA.
plugging these, we have:
=2sinC X 2cos ((A-B+C)/2) cos((A-B-C)/2).
=4 sinA sinB sin C QED.
hope its clear enough. have a good day:D!

Answer 4

dont know
all i know is my
+,-,times,divisions,mixed #s,fractions