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

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- July 4th 2008, 08:05 AM #2

- July 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

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

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