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Math Help - Help Me With This Trig ID proof Please Soon!

  1. #1
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    Help Me With This Trig ID proof Please Soon!

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  2. #2
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    Re: Help Me With This Trig ID proof Please Soon!

    Hey Melcarthus.

    Your equation is wrong. As an example let theta = 0, then on the LHS you get 0/0 for the answer but on the RHS you get 1/1.

    The equation is screwed up.
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  3. #3
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    Re: Help Me With This Trig ID proof Please Soon!

    Hello, Melcarthus!

    \text{Prove: }\:\frac{\sin\theta -\cos\theta + 1}{\sin\theta + \cos\theta - 1} \:=\:\frac{\sin\theta + 1}{\cos\theta}

    The left side is: . \frac{\sin\theta - (\cos\theta - 1)}{\sin\theta + (\cos\theta - 1)}

    Then: . \frac{\sin\theta - (\cos\theta-1)}{\sin\theta + (\cos\theta - 1)} \cdot \color{blue}{\frac{\sin\theta - (\cos\theta - 1)}{\sin\theta - (\cos\theta -1)}}

    . . . . . =\;\frac{\sin^2\!\theta  - 2\sin\theta(\cos\theta-1) + (\cos\theta-1)^2}{\sin^2\!\theta - (\cos\theta -1)^2}

    . . . . . =\;\frac{\sin^2\!\theta - 2\sin\theta\cos\theta + 2\sin\theta + \cos^2\!\theta - 2\cos\tjeta + 1}{\sin^2\!\theta - \cos^2\!\theta + 2\cos\theta - 1}

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

    . . . . . =\;\frac{2 + 2\sin\theta - 2\cos\theta - 2\sin\theta\cos\theta}{2\cos\theta - 2\cos^2\!\theta}

    . . . . . =\;\frac{2(1+\sin\theta) - 2\cos\theta(1 + \sin\theta)}{2\cos\theta(1-\cos\theta)}

    . . . . . =\;\frac{2(1+\sin\theta)(1-\cos\theta)}{2\cos\theta(1-\cos\theta)}

    . . . . . =\;\frac{1+\sin\theta}{\cos\theta}
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  4. #4
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    Re: Help Me With This Trig ID proof Please Soon!

    Hi Melcartus,
    a alternative solution:
    (sint-cost+1)*cost=(sint+cost-1)*(sint+1)
    You receive a identity, not forget, that sin^2(t)+cos^2(t)=1.
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