pls help me solv sin3A

weather

amit

pls help me solv sin3Ape pe pe pein

Salonii Sharma

sin 3A = 3 sin A - 4 sin^3 A

Right??

We ought to prove this. Now write Sin3A as

sin(2A + A)

apply the identity sin(A+B)=sinAcosb+cosAsinB

So you'll get sin2AcosA+cos2AsinA

Now sin 2A=2sinAcosA and cos2A is 1-2 sin^2A

You put these values and then solve. It's very easy.

However you don't get such questions. The type of question commonly asked are of sin5A and cos cos 5A and all. Because we already know the result of sin3A's and cos 3A's. You'll do the above in same manner.

:-)

edison

http://www.goiit.com/posts/list/trignometry-pls-help-me-prove-sin3a-1010504.htm

rahulmishra

That's really easy :-

Sin 3A = Sin (2A + A) = Sin 2A.Cos A + Cos 2A.Sin A
= (2 Sin A. Cos A) . Cos A + (1 - 2 Sin² A). Sin A
= 2 Sin A.Cos²A + Sin A - 2 Sin³ A
= 2 Sin A (1 - Sin²A) + Sin A - 2 Sin³A
= 2 Sin A - 2 Sin³ A + Sin A - 2 Sin³ A
= 3 Sin A - 4 Sin³ A.
And That's the answer

mayur khairnar

sin3A= sin(2A+A)

we know that sin(A+B)= sinAcosB+ cosAsinB

applying the same, we get

sin 3A= sin2AcosA+cos2AsinA

= 2sinAcosAcosA+( 1-2sin²A)sinA

= 2(1-sin²A)sinA+ sinA( 1-2sin²A)

=2sinA- 2sin³A+ sinA- 2sin³A = 3sinA- 4sin³A hence the solution

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