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Trig identity.Help

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    Need help in these two questions
    Thanks in advance
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    First part of first question from double angle identities, then it's a case of rearranging and factorising until your left with sinA/CosA = tanA
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    Second question square both sides to start and then rewrite the cos2x term and start working towards something that could resemble tanx
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    (Original post by mathcoachni)
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    First part of first question from double angle identities, then it's a case of rearranging and factorising until your left with sinA/CosA = tanA
    I did this part so far already and with one more step which is factorising Cos^2(A)-Sin^2(A) but got stuck after that
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    (Original post by lil_jack)
    I did this part so far already and with one more step which is factorising Cos^2(A)-Sin^2(A) but got stuck after
    Remember what your trying to make the LHS look like, you want sin/cos, sometimes it just takes you to go through the different options.
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    (Original post by lil_jack)
    I did this part so far already and with one more step which is factorising Cos^2(A)-Sin^2(A) but got stuck after that
    remember that

     cos(2A) \equiv cos^2A - sin^2A \equiv 2cos^2A - 1

    this is done by using  sin^2A + cos^2A \equiv 1 \implies sin^2A \equiv 1 - cos^2A

    now in the denominator, the -1 and +1 cancel out each other, so you are left with:

     \frac{2sinAcosA + sinA}{2cos^2A + cosA}

    Factorise this now and you should be able to get the answer
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    Thanks alot mate
 
 
 
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