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Math Help - TEST TOMORROW: Need help simplifying Trig Expressions

  1. #1
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    TEST TOMORROW: Need help simplifying Trig Expressions

    Sin x + 1- cos x
    1-cos x sin x


    cos x sec2x-cos x
    sin x tan x + cos x



    How would you simplify these two equations? (step-by-step)

    Thank you so much for help,
    Sean
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  2. #2
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    Re: TEST TOMORROW: Need help simplifying Trig Expressions

    Check the pdf, I hope it helps.

    In your first expression (not equation, expression), I'm not sure I understood what you wrote. The second one I'm pretty sure of.
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    Re: TEST TOMORROW: Need help simplifying Trig Expressions

    TEST TOMORROW: Need help simplifying Trig Expressions-08-apr-13.png
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    Re: TEST TOMORROW: Need help simplifying Trig Expressions

    Hello, srob22!

    \text{Simplify: }\:\frac{\sin x}{1-\cos x} + \frac{1-\cos x}{\sin x}

    Multiply the first fraction by \frac{1+\cos x}{1+\cos x}:

    . . \frac{\sin x}{1-\cos x}\cdot{\color{blue}\frac{1+\cos x}{1+\cos x}} + \frac{1-\cos x}{\sin x} \;=\;\frac{\sin x(1+\cos x)}{1-\cos^2\!x} + \frac{1-\cos x}{\sin x}

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

    . . =\;\frac{1+\cos x + 1 - \cos x}{\sin x} \;=\;\frac{2}{\sin x} \;=\;2\csc x




    \text{Simplify: }\:\frac{\cos x\sec^2\!x - \cos x}{\sin x\tan x +  \cos x}

    We have: . \dfrac{\cos x(\sec^2\!x - 1)}{\sin x\!\cdot\!\frac{\sin x}{\cos x} + \cos x} \;=\;\dfrac{\cos x\tan^2\!\!x}{\frac{\sin^2\!x}{\cos x} + \cos x} \;=\;\dfrac{\cos x\frac{\sin^2\!x}{\cos^2\!x}}{\frac{\sin^2\!x+\cos  ^2\!x}{\cos x}}

    . . . . . . . =\;\dfrac{\frac{\sin^2\!x}{\cos x}}{\frac{1}{\cos x}} \;=\; \dfrac{\sin^2\!x}{\cos x}\cdot\dfrac{\cos x}{1} \;=\;\sin^2\!x
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