Question

if A+B+C = π
find sin 2A+sin 2B+ sin 2C​

Answer 1

Answer:

sin2A + sin 2B +sin2C = 2 sin(A+B)cos(A-B) + 2sinC cosC =2sinC cos(A-B)+2sinC cosC =2sinC (cos(A-b) + cos C) =2sin C(cos(A-B) - cos(A+B) ) = 2sinC . 2sin A sin B =4 sinA sin B sin C

7

Answer 2

Answer:

$\maltese \: \bf \underline{{\sf4 \sin A \sin B \sin C}}$

Step-by-step explanation:

Given:

• A+B+C = π

To find:

sin2A + sin2B + sin2C

Solution:

$\sf\sin2A + \sin(2B ) + \sin(2C) \\ \\ :\implies \sf 2 \sin \bigg \lgroup \sf \frac{2A + 2B}{2} \bigg \rgroup\cos\bigg \lgroup \frac{2A - 2B}{2} \bigg \rgroup + \sin 2C$

$:\implies \sf2 \sin \: (A+B) \cos(A - B) + \sin2C \\ \\ :\implies \sf2 \sin \: (\pi -C ) \cos(A - B) + \sin2C$

Therefore,

$\bigg \lgroup\tt \: A+B+C = π, \: \: A+B = \pi - C \\ \therefore \sf \sin(A+B) = \sin(\pi - C ) = \sin \: C \bigg \rgroup$

$: \implies \sf2 \sin \: C \cos(A - B) + 2\sin \: C \cos \: C \\ \\ : \implies \sf2 \sin \: C \big \lgroup\cos(A - B) + cos \: C ) \big \rgroup \\ \\ \implies \sf2 \sin \: C \big \lgroup \cos(A - B) - \cos(A - B)\big \rgroup$

$\star\therefore \sf \cos(A - B) - \cos(A+B) = 2 \sin \: A \sin B$

$: \implies \sf \: 2 \sin C[2 \sin \sin \: B ] \\ \\ : \implies \bf\sf4 \sin A \sin B \sin C$