hii friend help me in this question for friends
Attachments:
Answers
Answered by
1
sin 3A=sin (2A + A)
= sin 2A cos A + cos 2A sin A
= 2 sin A cos A ∙ cos A + (1 - 2 sin^2 A) sin A
= 2 sin A (1 - sin^2 A) + sin A - 2 sin^3 A
= 2 sin A - 2 sin^3 A + sin A - 2 sin^3 A
= 3 sin A - 4 sin^3 A
Therefore, sin 3A = 3 sin A - 4 sin^3 A Proved
= sin 2A cos A + cos 2A sin A
= 2 sin A cos A ∙ cos A + (1 - 2 sin^2 A) sin A
= 2 sin A (1 - sin^2 A) + sin A - 2 sin^3 A
= 2 sin A - 2 sin^3 A + sin A - 2 sin^3 A
= 3 sin A - 4 sin^3 A
Therefore, sin 3A = 3 sin A - 4 sin^3 A Proved
pragya80:
ok
I'm not getting ur question
i think that its wrong
plz check ur question
Similar questions