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Math Help - trig identity

  1. #1
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    trig identity



    i need help with this identity, could someone tell me where i went wrong & help me?
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  2. #2
    Super Member

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    Hello, boonies!

    Your simplifying was off . . .


    ({\color{blue}\tan\theta\cos\theta} + 2\cos\theta)^2 + \left(2\sin\theta - {\color{red}\frac{\sin\theta}{\tan\theta}}\right)^  2 \;=\;5
    . . {\color{blue}\tan\theta\cos\theta \:=\:\frac{\sin\theta}{\cos\theta}\!\cdot\!\cos\th  eta \:=\:\sin\theta}

    . . {\color{red}\frac{\sin\theta}{\tan\theta} \:=\:\frac{\sin\theta}{\frac{\sin\theta}{\cos\thet  a}} \:=\:\cos\theta}


    The problem becomes: . ({\color{blue}\sin\theta} + 2\cos\theta)^2 + (2\sin\theta - {\color{red}\cos\theta})^2

    . . = \;\sin^2\!\theta + 4\sin\theta\cos\theta + 4\cos^2\!\theta + 4\sin^2\!\theta - 4\sin\theta\cos\theta + \cos^2\!\theta

    . . = \;5\sin^2\!\theta + 5\cos^2\!\theta \;=\;5\underbrace{(\sin^2\!\theta + \cos^2\!\theta)}_{\text{This is 1}} \;=\;5(1) \;=\;5

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