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

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

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

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

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

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

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