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Math Help - Trig Expressions

  1. #1
    Newbie Tom G's Avatar
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    Trig Expressions

    Hi, I've got to re-write 3sin x cos x + 4cos^2 x in the form a sin2x+b sin2x +c and I'm struggling. I've already asked my maths teacher for help and she told me how to do it but I can only remember half of the process and am now confused. Any help would be appreciated.
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  2. #2
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    Hello, Tom G!

    I've got to re-write 3\sin x\cos x + 4\cos^2\!x
    in the form a\sin2x + b\sin2x +c .
    . . . This can't be right!
    . . Do you mean: . {\color{blue}a\sin^2\!x + b\sin2x + c}
    You're expected to know these two identities: . \begin{array}{ccc}2\sin\theta\cos\theta & = & \sin2\theta\\<br />
\cos^2\!\theta & = & 1 - \sin^2\!\theta\end{array}

    Then: . 3\sin x\cos x + 4\cos^2\!x \;=\;\frac{3}{2}(2\sin x\cos x) + 4\left(1 - \sin^2\!x\right)

    . . = \;\frac{3}{2}\sin2x + 4 - 4\sin^2\!x \;=\; -4\sin^2\!x + \frac{3}{2}\sin2x + 4

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  3. #3
    Newbie Tom G's Avatar
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    Hi, I've checked the wording of the answer and have copied it down exactly:
    "Express 3sin x cos x + 4sin^2 x in the form a sin2x +b cos2x +c where a,b and c are constants to be found." - sorry i copied it down wrong.
    Last edited by Tom G; October 3rd 2007 at 12:29 PM. Reason: put the wrong question in
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  4. #4
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    Hello, Tom G!

    Okay, that makes it clearer . . .


    Express . 3\sin x\cos x + 4\sin^2\!x .in the form . a\sin2x + b\cos2x + c
    We will use: . 2\sin x\cos x \:=\: \sin2x\qquad \sin^2\!x \:=\: \frac{1 - \cos2x}{2}

    We have: . 3\sin x\cos x + 4\sin^2x \;=\;\frac{3}{2}(2\sin x\cos x) + 4\left(\frac{1 - \cos2x}{2}\right)

    . . = \;\frac{3}{2}\sin2x + 2(1 - \cos2x) \;=\;\frac{3}{2}\sin2x - 2\cos2x + 2

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