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Thread: Trig identities!!

  1. #1
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    Trig identities!!

    Hi all, Im stumped on this one:

    (sin s)/(1 + cos s) + (1 + cos s)/(sins s) = 2csc s

    Thanks for you help.
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  2. #2
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    Quote Originally Posted by xxpl510xx View Post
    Hi all, Im stumped on this one:

    (sin s)/(1 + cos s) + (1 + cos s)/(sins s) = 2csc s

    Thanks for you help.
    $\displaystyle \frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}$

    $\displaystyle \frac{(\sin x)(\sin x)+(1+\cos x)(1+\cos x)}{\sin x(1+\cos x)}$

    $\displaystyle \frac{\sin^2 x+1+2\cos x+\cos^2 x}{\sin x(1+\cos x)}$

    $\displaystyle \frac{1+1+2\cos x}{\sin x(1+\cos x)}$

    $\displaystyle \frac{2(1+\cos x)}{\sin x(1+\cos x)}$

    $\displaystyle \frac{2}{\sin x}$

    $\displaystyle 2\csc x$
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  3. #3
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    Hello, xxpl510xx!

    Another approach . . .


    $\displaystyle \frac{\sin \theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} \:= \:2\csc\theta$

    Multiply the first fraction by $\displaystyle \frac{1-\cos\theta}{1 - \cos\theta}$

    . . $\displaystyle \frac{\sin\theta}{1 + \cos\theta}\cdot\frac{1 - \cos\theta}{1 - \cos\theta} \:=\:\frac{\sin\theta(1 - \cos\theta)}{1-\cos^2\theta} \:=\:\frac{\sin\theta(1 - \cos\theta)}{\sin^2\theta} $ $\displaystyle =\:\frac{1 - \cos\theta}{\sin\theta}$


    And the problem becomes: .$\displaystyle \frac{1-\cos\theta}{\sin\theta} + \frac{1 + \cos\theta}{\sin\theta} \:=\:\frac{2}{\sin\theta} \:=\:2\csc\theta$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    This is a popular "trick" to use when dealing with:
    . . . $\displaystyle 1 + \sin\theta,\;1 - \sin\theta,\;1 + \cos\theta,$ or $\displaystyle 1 - \cos\theta$

    It arises from: .$\displaystyle \sin^2\theta + \cos^2\theta\:=\:1\quad\Rightarrow\quad \sin^2\theta \:=\:1-\cos^2\theta$

    . . $\displaystyle \Rightarrow\quad(\sin\theta)(\sin\theta) \:=\:(1-\cos\theta)(1 + \cos\theta)$

    which produces variations like: .$\displaystyle \frac{\sin\theta}{1 + \cos\theta} \:=\:\frac{1 - \cos\theta}{\sin\theta} $


    It also applies to: .$\displaystyle \sec\theta +1$ and $\displaystyle \sec\theta - 1$


    Just one more thing to watch for . . .

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