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Math Help - Sin/Cos Identity

  1. #1
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    Sin/Cos Identity

    How would i prove this identity
    (cos2θ)/(1+sin2θ) = (cotθ-1)/(cotθ+1)
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  2. #2
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    Hello, DjamaalRobinson!

    Prove: .  \frac{\cos2\theta}{1+\sin2\theta} \;=\;\frac{\cot\theta - 1}{\cot\theta + 1}

    The right side is: . \frac{\dfrac{\cos\theta}{\sin\theta} - 1}{\dfrac{\cos\theta}{\sin\theta} + 1}


    Multiply by \frac{\sin\theta}{\sin\theta}\!:\;\;\frac{\sin\the  ta\left(\dfrac{\cos\theta}{\sin\theta} - 1\right)}{\sin\theta\left(\dfrac{\cos\theta}{\sin\  theta} + 1\right)} \;=\;\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}


    Multiply by \frac{\cos\theta + \sin\theta}{\cos\theta + \sin\theta}\!: . \frac{\cos\theta+\sin\theta}{\cos\theta+\sin\theta  } \cdot\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} . =\;\frac{\cos^2\!\theta - \sin^2\!\theta}{\cos^2\!\theta + 2\sin\theta\cos\theta + \sin^2\!\theta}


    We have: . \frac{\overbrace{\cos^2\!\theta - \sin^2\!\theta}^{\text{This is }\cos2\theta}}{\underbrace{\sin^2\!\theta + \cos^2\!\theta}_{\text{This is 1}} + \underbrace{2\sin\theta\cos\theta}_{\text{This is }\sin2\theta}} \;=\;\frac{\cos2\theta}{1 + \sin2\theta} . . . ta-DAA!

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  3. #3
    Math Engineering Student
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    \begin{aligned}<br />
   \frac{\cos 2\theta }{1+\sin 2\theta }&=\frac{(\cos \theta +\sin \theta )(\cos \theta -\sin \theta )}{1+2\sin \theta \cos \theta } \\ <br />
 & =\frac{(\cos \theta +\sin \theta )(\cos \theta -\sin \theta )}{(\cos \theta +\sin \theta )^{2}} \\ <br />
 & =\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=\frac{\cot \theta -1}{\cot \theta +1}.\quad\blacksquare<br />
\end{aligned}
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