Question

When
A + B + C = pi
prove that
Sin 2A + sin 2B - sin 2C = 4cosAcosBcosC
hence deduced that
sin 2A + sin2B +sin 2C = 4sinAsinBsinC

Answer 1

Pls guys !!! Help me to solve this one!

Answer 2

Use two product-to-sum formulas.
\[sinAsinB = \frac{ 1 }{ 2 } [\cos(A-B) - \cos(A+B)]\]
and
\[cosAsinB = \frac{ 1 }{ 2 } [\sin(A+b) - \sin (A-B)]\]

Answer 3

Let's try.
\[4sinAsinBsinC = 4(\frac{ 1 }{ 2 } [\cos(A-B)-\cos(A+B)]) sinC\]
\[=2\cos(A-B)sinC-2\cos(A+B)sinC\]
\[2(\frac{ 1 }{ 2 }[\sin(A-B+C)-\sin(A-B-C)]) - 2(\frac{ 1 }{ 2}[\sin(A+B+C)-\sin(A+B-C)])\]
\[=\sin(A-B+C)-\sin(A-B-C)-\sin(A+B+C)+\sin(A+B-C)\]