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Math Help - [Urgent - Trig] How would I prove the followin equation with double angle formulas?

  1. #1
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    [Urgent - Trig] How would I prove the followin equation with double angle formulas?

    cotx = csc2x + cot2x

    My Attempt:

    RS:
    = 1/sin2x + sin2x/cos2x
    = (cos2x + sin2x) / (sin2x)(cos2x)
    = (2cos^2x - 1 + 2sinxcosx) / (2sinxcosx + 1 - 2sin^2x)
    = 2cos^2x - 1 / 1 - 2sin^2x
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Macleef View Post
    cotx = csc2x + cot2x

    My Attempt:

    RS:
    = 1/sin2x + sin2x/cos2x
    = (cos2x + sin2x) / (sin2x)(cos2x)
    = (2cos^2x - 1 + 2sinxcosx) / (2sinxcosx + 1 - 2sin^2x)
    = 2cos^2x - 1 / 1 - 2sin^2x
    ALWAYS write everything in terms of sin and cos

    \frac{cos x}{sin x} = \frac{1}{sin 2x} + \frac{cos 2x}{sin 2x}

    RHS:

    = \frac{1 + cos 2x}{sin 2x}

    = \frac{1 + ( 2cos^2 x - 1)}{2 \ Sin x \ Cos x}

    = \frac{2cos^2 x}{2 \ sin x \ Cos x}

    = \frac{cos x}{sin x}

    = LHS

    \rightarrow \mathfrak{Quad \ Erat \ Demonstrandum}
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  3. #3
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    Hello, Macleef!

    By now, you've seen your mistake . . .


    Prove: . \cot x \:= \:\csc2x + \cot2x

    My Attempt:
    RS \:=\:\frac{1}{\sin2x} + {\color{red}\frac{\sin2x}{\cos2x}} . ← Here!

    We have: . \frac{1}{\sin2x} + \frac{\cos2x}{\sin2x} \;= \;\frac{1 + \cos2x}{\sin2x} \;=\;\frac{2\cos^2x}{2\sin x\cos x} \;=\;\frac{\cos x}{\sin x} \;=\;\cot x<br /> <br />



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