Hi Guys

I am a bit stuck on a question please could anyone help?

it is 'By expressing sin 3A as sin (2A+A), find an expression for sin 3A in terms of Sin A.'

thankyou

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- Jul 4th 2008, 07:49 AM #1

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- Jul 4th 2008, 08:05 AM #2
Hello

The expression $\displaystyle \sin(3A)=\sin(2A+A)$ can be expanded using $\displaystyle \sin(x+y)=\sin x\cos y+\sin y\cos x$ with $\displaystyle x=2A$ and $\displaystyle y=A$. It should give you another expression which you'll have to transform using the three following identities :

$\displaystyle \begin{cases}\sin (2A)=2\sin A\cos A\\ \cos^2A=1-\sin ^2A\\\cos (2A)=1-2\sin^2A\end{cases}$

Does it help ?

- Jul 4th 2008, 07:46 PM #3

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This is good for practice.

sin(3A)

= sin(2A +A)

= sin(2A)cosA +cos(2A)sinA

= [2sinAcosA]cosA +(cos^2(A) -sin^2(A)]sinA

= 2sinAcos^2(A) +sinAcos^2(A) -sin^3(A)

= 3sinAcos^2(A) -sin^3(A)

= 3sinA[1 -sin^2(A)] -sin^3(A)

= 3sinA -3sin^3(A) -sin^3(A)

= 3sinA -4sin^3(A) -----------answer.

I used:

sin(2X) = 2sinXcosX

cos(2X) = cos^2(X) -sin^2(X)

sin^2(X) +cos^2(X) = 1

- Jul 5th 2008, 05:20 AM #4

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- Dec 2007
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