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Thread: Simplgying trig functions

  1. #1
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    Simplgying trig functions

    Simplify the following:


    (sin q + tan q) / (1 + sec q)
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  2. #2
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    Quote Originally Posted by Mr_Green View Post
    Simplify the following:


    (sin q + tan q) / (1 + sec q)
    Multiply numerator and denominator by,
    $\displaystyle \cos x$
    Thus,
    $\displaystyle \frac{\sin x\cos x+\tan x \cos x}{\cos x+\sec x\cos x}$
    But,
    $\displaystyle \tan x\cos x=\frac{\sin x}{\cos x}\cdot \cos x=\sin x$
    And,
    $\displaystyle \sec x\cos x=1$
    Thus,
    $\displaystyle \frac{\sin x\cos x+\sin x}{\cos x+1}$
    Factor,
    $\displaystyle \frac{\sin x(\cos x+1)}{(\cos x+1)}=\sin x$
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  3. #3
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    Hello, Mr_Green!

    Simplify: .$\displaystyle \frac{\sin\theta + \tan\theta}{1 + \sec\theta}$

    We have: .$\displaystyle \frac{\sin\theta + \frac{\sin\theta}{\cos\theta}}{1 + \frac{1}{\cos\theta}} $


    Multiply by $\displaystyle \frac{\cos\theta}{\cos\theta}\!:\;\;\frac{\cos\the ta}{\cos\theta}\cdot\frac{\sin\theta + \frac{\sin\theta}{\cos\theta}}{1 + \frac{1}{\cos\theta}} \;= \;\frac{\sin\theta\cos\theta + \sin\theta}{\cos\theta + 1} $


    Factor: .$\displaystyle \frac{\sin\theta(\cos\theta + 1)}{\cos\theta + 1} \;=\;\sin\theta$

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