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Thread: Can someone help me solve this identity? :)

  1. #1
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    Can someone help me solve this identity? :)



    I've tried many things, can't seem to work it out. Thank you!
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  2. #2
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    Krizalid's Avatar
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    Did you tried simplifyin' the LHS?

    Factorise.
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  3. #3
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    Yes, I've been working with the left hand side.

    I've tried a lot of things.
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  4. #4
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    Krizalid's Avatar
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    $\displaystyle \frac{{\tan \theta - \sin \theta }}
    {{\sin ^3 \theta }} = \frac{{\sin \theta \left( {\dfrac{1}
    {{\cos \theta }} - 1} \right)}}
    {{\sin ^3 \theta }} = \frac{{1 - \cos \theta }}
    {{\cos \theta (1 + \cos \theta )(1 - \cos \theta )}}.$

    Does that make sense?
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  5. #5
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    How does sin^3theta = (1+costheta)(1-costheta)?
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  6. #6
    Super Member angel.white's Avatar
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    $\displaystyle \frac{tan\theta - sin\theta}{sin^{3}\theta}$

    1. because $\displaystyle tan\theta = \frac{sin\theta}{cos\theta}$, substitute that into $\displaystyle tan\theta$
    $\displaystyle \frac{\frac{sin\theta}{cos\theta} - sin\theta}{sin^{3}\theta}$

    2. Common denominator of the fractions in the numerator
    $\displaystyle \frac{\frac{sin\theta}{cos\theta} - \frac{sin\theta cos\theta}{cos\theta}}{sin^{3}\theta}$

    3. combine fractions in the numerator
    $\displaystyle \frac{\frac{sin\theta-sin\theta cos\theta}{cos\theta}}{sin^{3}\theta}$

    4. simplify (dividing a fraction by something means multiplying by it's inverse):
    $\displaystyle \frac{sin\theta-sin\theta cos\theta}{cos\theta sin^{3}\theta}$

    5. factor out $\displaystyle sin\theta$ in the numerator
    $\displaystyle \frac{sin\theta (1 - cos\theta)}{cos\theta sin^{3}\theta}$

    $\displaystyle sin\theta$ in the numerator and denominator cancel eachother out
    $\displaystyle \frac{1 - cos\theta}{cos\theta sin^{2}\theta}$

    6. Multiply the numerator and the denominator by $\displaystyle 1+cos\theta$
    $\displaystyle \frac{(1 - cos\theta)(1 + cos\theta)}{cos\theta sin^{2}\theta (1 + cos\theta)}$

    7. FOIL the numerator:
    $\displaystyle \frac{1 + cos^{2}\theta}{(1 + cos\theta)cos\theta sin^{2}\theta}$

    8. Using the formula $\displaystyle sin^{2}\theta + cos^{2}\theta = 1$ see that $\displaystyle sin^{2}\theta = 1 - cos^{2}\theta$ and substitute that into the equation

    $\displaystyle \frac{1 - cos^{2}\theta}{(1 + cos\theta)cos\theta (1- cos^{2}\theta)}$

    9. $\displaystyle 1- cos^{2}\theta$ is in the numerator and the denominator, so cancel them out
    $\displaystyle \frac{1}{(1 + cos\theta)cos\theta}$
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  7. #7
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    Hello, Jeavus

    How does sin^3theta = (1+costheta)(1-costheta)?
    It doesn't . . .


    Quote Originally Posted by Krizalid

    $\displaystyle \frac{{\tan \theta - \sin \theta }}{{\sin ^3 \theta }}$

    . . $\displaystyle = \;\frac{{\sin \theta \left( {\dfrac{1}{{\cos \theta }} - 1} \right)}}{{\sin ^3 \theta }}$ . . . . cancel a $\displaystyle {\color{blue}\sin\theta}$

    . . $\displaystyle {\color{blue}= \;\dfrac{\dfrac{1}{\cos\theta} - 1}{\sin^2\theta} \;=\;\dfrac{\dfrac{1-\cos\theta}{\cos\theta}}{\sin^2\theta} \;=\;\frac{1-\cos\theta}{\cos\theta\sin^2\theta} \;=\;\frac{1-\cos\theta}{\cos\theta(1-\cos^2\theta)}}$

    . . $\displaystyle =\; \frac{{1 - \cos \theta }}
    {{\cos \theta (1 + \cos \theta )(1 - \cos \theta )}}$
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  8. #8
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    Thank you all so much.
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