Question

If A+B+C=180° , prove that sin²A+sin²B+sin²C=2(1+cosAcosBcosC) give me the answer fastly ​

Answer 1

$\huge{\mathfrak{\underline{\red{Answer!}}}}$

Given,

A+B+C=180° , prove sin²A+sin²B+sin²C=2(1+cosAcosBcosC)

Let's LHS = sin2A + sin2B + sin2C

so that

2S = 2sin2A + 1 – cos2B +1 – cos2C

= 2 sin2A + 2 – 2cos(B + C)

cos(B – C)

= 2 – 2 cos2A + 2

– 2cos(B + C) cos(B – C)

∴ S = 2 + cosA [cos(B – C) + cos(B+

C)] since cosA

= – cos(B+C)

∴ S = 2 + 2 cos A cos B cos C

= 2(1+ cos A cos B cos C)

= RHS

Therefore, LHS = RHS is

proved.

Answer 2

Answer:

trigonometry yeh.......