We Know
Sin (A + B) = sin A cos B + cos A sin B
Putting B = 4A we get
sin 5A = sin ( A + 4 A)
= sin A cos 4A + cos A sin 4A
= sin A (1- 2 Sin^2 2A) + cos A 2. Sin 2A cos 2A
( putting cos 2x = 1 – 2 sin ^2 x and sin 2x = 2 sin x cos x and x = 2A)
= sin A(1- 8 Sin^2 A cos^2 A) + Cos A 4 Cos A sin A (1- 2 Sin^2 A)
(putting cos 2x = 1 – 2 sin ^2 x and sin 2x = 2 sin x cos x and x = A)
= sin A - 8 sin ^3A (1-sin^2 A) + 4 sin A Cos ^2 A(1- 2 sin ^2 A)
= sin A - 8 sin ^3 A + 8 Sin ^5 A + 4 sin A(1-sin^2A)(1-2 Sin ^2 A)
= sin A - 8 sin ^3 A + 8 sin ^5 A + 4 sin A ( 1- 3 sin ^2A + 2 Sin ^4 A)
= 5 SIn A - 20 sin ^3 A + 16 sin ^5 A
(note: edited the above based on 3rd comment below to keep the flow)
7 comments:
You are interested in mathematics? Can't believe it. I never understood all these trig proofs or the need for them. Was glad when I left mathematics behind.
Good effort anyway. Might be useful for someone.
Sorry about the above comment. I think it was not appropriate. Always hated trig identities `and their endless proofs. Mathematics in general caused me endless trouble. Hence the outburst.
Nice blog you have there. May be you can have a writeup on how to solve these type of problems in trig. That will be of great help to a lot of people who struggle like I did.
when
sin A - 8 sin ^3 A + 8 sin ^5 A + 4 sin A ( 1- 3 sin ^2A + 2 Sin ^5 A)
you got wrong on (1 - 3Sin^2A + 2Sin^5A)
the right answer is (1- 3Sin^2A + 2Sin^4A)
then you multiply with 4 sinA = 4Sin A - 12 Sin ^3 A + 8 Sin ^5A
am i Right?
thanks Loray. I incorporated the comment to keep the flow.
Could you please me...how you found..
sinA(1-2sin^2A) to sinA(1-8sin^2Acos^2A)
if cosx=1/2(a+1/a) then prove that cos2x=1/2(a^2+1/a^2)
Thanks for the query. It was a typo it should be
sin A ( 1- 2 sin^2 2A)
I have done the correction to keep the flow
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