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Thread: Trig identities

  1. #1
    dolphinlover527
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    Unhappy Trig identities

    1-{(sin^2)/(1+cos x)}=cos x
    How?
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  2. #2
    Senior Member topher0805's Avatar
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    Is this what you mean?

    Prove that,

    $\displaystyle 1 - \frac{sin^2 x}{1 + cos x} = cos x$

    If so, then it is not as hard as you think. Everything is already changed to $\displaystyle sine$ and $\displaystyle cosine$, so half of the work is done.

    First, multiply both sides by $\displaystyle 1 + cos x$:

    $\displaystyle 1 + cos x - sin^2 x = cos x + cos^2 x$

    Then subtract cos x from both sides:

    $\displaystyle
    1 - sin^2 x = cos^2 x$

    Add $\displaystyle sin^2 x $to both sides and we now have that:

    $\displaystyle cos^2 x + sin^2 x = 1$

    This is a trigonometric identity, and therefore you have proved that:

    $\displaystyle 1 - \frac{sin^2 x}{1 + cos x} = cos x$
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  3. #3
    Super Member

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    Hello, dolphinlover527!

    Prove: .$\displaystyle 1-\frac{\sin^2\!x}{1+\cos x} \:=\:\cos x$

    We have: .$\displaystyle 1 - \frac{\sin^2\!x}{1+\cos x} \;\;=\;\;1 - \frac{1-\cos^2\!x}{1+\cos x} \;\;=\;\;1 - \frac{(1-\cos x)(1+\cos x)}{1+\cos x} $


    Reduce: .$\displaystyle 1 - (1 - \cos x) \;\;=\;\;1 - 1 + \cos x \;\;=\;\;\cos x$

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