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    Hi

    looking in syllabus and we need to know how this works:

    sin3x = sin(2x+x) = sinx(3-4sin^2(x))

    Can anyone please explain this to me??

    Thanks in advance!
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    sin3x = sin(2x+x)
    = sin2xcosx + cos2xsinx
    = 2sinxcosxcosx + (cos²x - sin²x)sinx
    = sinx (2cos²x + cos²x - sin²x)
    = sinx (3cos²x - sin²x)

    then use:
    cos²x = 1-sin²x

    =sinx (3[1-sin²x]-sin2x)
    =sinx (3-4sin²x)
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    (Original post by alwaysingtogeth)
    Hi

    looking in syllabus and we need to know how this works:

    sin3x = sin(2x+x) = sinx(3-4sin^2(x))

    Can anyone please explain this to me??

    Thanks in advance!
    \sin 3x=\sin (2x+x)\\

=\sin 2x \cos x + \cos 2x \sin x\\

=2\sin x \cos ^{2} x + \sin x (1-2\sin ^{2} x)\\

=2\sin x (1-\sin ^{2} x) + \sin x -2\sin ^{3} x\\

=2\sin x - 2\sin ^{3} x + \sin x - 2\sin^{3} x\\

=3\sin x - 4\sin ^{3} x\\

=\sin x(3-4\sin ^{2} x)
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    Cheers people! Had a question similar in exam today

    Prove cos3x = something and I managed to get it right.

    Thanks again
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    (Original post by alwaysingtogeth)
    Cheers people! Had a question similar in exam today

    Prove cos3x = something and I managed to get it right.

    Thanks again
    no problem. anytime
 
 
 
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