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    I was doing C4 solomon paper C Q6C and i have to integrate: sin^2 (2t)

    And in the markscheme it says
    sin^2 (2t) = 4sin^2 (t) cos^2 (t)

    Can anyone explain where they got that from?
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    (Original post by Economistician)
    I was doing C4 solomon paper C Q6C and i have to integrate: sin^2 (2t)

    And in the markscheme it says
    sin^2 (2t) = 4sin^2 (t) cos^2 (t)

    Can anyone explain where they got that from?
    Use the half angle formula for sin(2t).
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    (Original post by Economistician)
    I was doing C4 solomon paper C Q6C and i have to integrate: sin^2 (2t)

    And in the markscheme it says
    sin^2 (2t) = 4sin^2 (t) cos^2 (t)

    Can anyone explain where they got that from?
    You should know that \sin 2t = 2 \sin t\cos t, so squaring both sides yields:

    \displaystyle 

\begin{equation*}\sin^2 t = (2\sin t\cos t)^2 = 2^2 \sin^2 t \cos^2 t = \cdots \end{equation*}

    Edit: ninja-ed.
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    (Original post by Zacken)
    You should know that \sin 2t = 2 \sin t\cos t, so squaring both sides yields:

    \displaystyle 

\begin{equation*}\sin^2 t = (2\sin t\cos t)^2 = 2^2 \sin^2 t \cos^2 t = \cdots \end{equation*}

    Edit: ninja-ed.
    You've put more effort, more likely he'll understand what you've said.
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    Is it as simple as:
    Sin2x = 2SinxCosx
    Sin^2 (2x) = 4Sin^2(x)Cos^2(x) ?

    Edit: Oh, ok thanks for your help. I dont know why that had me stumped for so long
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    (Original post by Economistician)
    Is it as simple as:
    Sin2x = 2SinxCosx
    Sin^2 (2x) = 4Sin^2(x)Cos^2(x) ?

    Edit: Oh, ok thanks for your help. I dont know why that had me stumped for so long
    Yep.
 
 
 
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