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    Hi quick question.

    On D do you rearrange cos squared theta + sin squared theta = 1 to cos squared theta - 1 = sin squared theta to remove the sin squared theta from the question? I'm a little confused with it.

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     2\sin^2 \theta -2\cos\theta - \cos^2\theta = 1

    \rightarrow 2(1- \cos^2\theta) -2\cos\theta -\cos^2\theta = 1

    then solve as quadratic.
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    (Original post by ThrashMetal)
    Hi quick question.

    On D do you rearrange cos squared theta + sin squared theta = 1 to cos squared theta - 1 = sin squared theta
    That is not correct

    \cos ^2 \theta + \sin ^2 \theta = 1

    Subtract \cos ^2 \theta from both sides of the equation
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    (Original post by TenOfThem)
    That is not correct

    \cos ^2 \theta + \sin ^2 \theta = 1

    Subtract \cos ^2 \theta from both sides of the equation
    So Sin squared theta = 1-Cos squared theta? Does that mean I can now remove a sin squared theta from the original equation leaving me with 2-2 cos theta - cos squared theta = 1? I'm so confused and I don't know why.
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    (Original post by ThrashMetal)
    So Sin squared theta = 1-Cos squared theta? Does that mean I can now remove a sin squared theta from the original equation leaving me with 2-2 cos theta - cos squared theta = 1? I'm so confused and I don't know why.
    See post 2
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    (Original post by TenOfThem)
    See post 2
    Woops my bad. So basically whatever you rearrange cos^2theta+sin^2theta=1 for, you remove the = X and put brackets around what comes before the equals sign, where the = X was?
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    (Original post by ThrashMetal)
    Woops my bad. So basically whatever you rearrange cos^2theta+sin^2theta=1 for, you remove the = X and put brackets around what comes before the equals sign, where the = X was?
    erm

    I really have no idea what you mean by this

    \sin ^2 \theta = 1 - \cos ^2 \theta

    So

    2 \sin ^2 \theta = 2(1 - \cos ^2 \theta)
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    (Original post by TenOfThem)
    erm

    I really have no idea what you mean by this

    \sin ^2 \theta = 1 - \cos ^2 \theta

    So

    2 \sin ^2 \theta = 2(1 - \cos ^2 \theta)
    Yeah I had trouble explaining what I meant. Basically you can remove what the equation equals to and put what ever is left in brackets and solve from there is what I understand.

    Thanks for the help guys, I think I fully understand it now
 
 
 
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