Question

evaluate sin 5 A trigonometry

Answer 1

sin5A=
sin(2A+3A)=sin2A.cos3A+cos2A.sin3A
...by formula of sin(A+ B)
or we hv formula of sin2A=2sinA.cosB
so,sin5A=2sin[5A÷2].cos[5A÷2]

Answer 2

Hello Friend,

The solution is in the picture.
The identities used are:

• sin (A+B) = sin A cos B + cos A sin B
• sin 2A = 2 sin A cos A
• sin 3A = 3 sin A - 4 sin³A
• cos 2A = 1 - 2 sin²A
• cos 3A = 4 cos³A - 3 cos A
• cos²A = 1 - sin²A


Using only this much, sin 5A can be evaluated in terms of sin A.

Hope it helps.

Purva
@Purvaparmar1405
Brainly.in