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double angle formula

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I changed sin1/2 x to sin(x-1/2x) but this didnt seem to lead anywhere im really stuck
this might help its the reverse but it will give you a hint
http://www.mathguide.com/lessons2/HAF.html
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fiftythree
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Original post by Zainab88
this might help its the reverse but it will give you a hint
http://www.mathguide.com/lessons2/HAF.html

I know how to derive it from cos2x now but how do i relate this to the question
Original post by fiftythree
I know how to derive it from cos2x now but how do i relate this to the question


Rearrange cos⁡2x=1−2sin⁡2x\cos 2x = 1-2\sin^2 x for sin⁡x\sin x.

And then replace all xx with x2\dfrac{x}{2}.
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fiftythree
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Original post by RDKGames
Rearrange cos⁡2x=1−2sin⁡2x\cos 2x = 1-2\sin^2 x for sin⁡x\sin x.

And then replace all xx with x2\dfrac{x}{2}.

Thanks for the reply

I rearranged it getting sinx=+/-Square root of 1/2-cos2x

When i sub x/2 for x i dont get the right answer
Original post by fiftythree
Thanks for the reply

I rearranged it getting sinx=+/-Square root of 1/2-cos2x

When i sub x/2 for x i dont get the right answer


Rearranging, you should get

sin⁡x=±1−cos⁡2x2\sin x = \pm \sqrt{\dfrac{1-\cos 2x}{2}}

and NOT

sin⁡x=±12−cos⁡2x\sin x = \pm \sqrt{\dfrac{1}{2} - \cos 2x}

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