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Thread: Simplify Expression

  1. #1
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    Simplify Expression

    sin x / 1 + cos x + cot x

    sin^2 x / 1 - cos x

    sec^2 x -1 / sec^2 x

    Thanks!
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  2. #2
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    Hello,

    Quote Originally Posted by NeedHelp18 View Post
    sin x / 1 + cos x + cot x

    sin^2 x / 1 - cos x

    sec^2 x -1 / sec^2 x

    Thanks!
    Please put parenthesis...

    I guess the second one is :

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

    We know that $\displaystyle \sin^2 x+\cos^2 x=1 \implies \sin^2 x=1-\cos^2 x=(1-\cos x) \cdot (1+\cos x)$

    So, how does it simplify ?

    (not sure about this one)
    $\displaystyle \frac{\sec^2 x-1}{\sec^2 x}$ (or $\displaystyle \sec^2 x-\frac{1}{\sec^2 x}$ ?)

    $\displaystyle =1-\frac{1}{\sec^2 x}=1-\cos^2 x=...$

    Is the first one $\displaystyle \frac{\sin x}{1+\cos x}+\cot x$ or $\displaystyle \frac{\sin x}{1+\cos x+\cot x}$ ?
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,



    Please put parenthesis...

    I guess the second one is :

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

    We know that $\displaystyle \sin^2 x+\cos^2 x=1 \implies \sin^2 x=1-\cos^2 x=(1-\cos x) \cdot (1+\cos x)$

    So, how does it simplify ?

    (not sure about this one)
    $\displaystyle \frac{\sec^2 x-1}{\sec^2 x}$ (or $\displaystyle \sec^2 x-\frac{1}{\sec^2 x}$ ?)

    $\displaystyle =1-\frac{1}{\sec^2 x}=1-\cos^2 x=...$

    Is the first one $\displaystyle \frac{\sin x}{1+\cos x}+\cot x$ or $\displaystyle \frac{\sin x}{1+\cos x+\cot x}$ ?
    the first option
    thanks
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    Quote Originally Posted by NeedHelp18 View Post
    the first option
    thanks
    For the third one ?

    $\displaystyle \frac{\sin x}{1+\cos x}+\cot x=\frac{\sin x}{1+\cos x}+\frac{\cos x}{\sin x}$

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

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

    $\displaystyle =\frac{\cos x+1}{\sin x(1+\cos x)}=\dots$
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  5. #5
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    is the answer csc x
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  6. #6
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    ( in this one arent u supposed to leave the sinx in the numerator)



    i did it again and now i get sin x + cosx as the answer.
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  7. #7
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    Quote Originally Posted by NeedHelp18 View Post
    ( in this one arent u supposed to leave the sinx in the numerator)



    i did it again and now i get sin x + cosx as the answer.
    $\displaystyle =\frac{\color{red}\cos x+1}{\sin x({\color{red}1+\cos x})}$

    Can you just simplify ?



    The answer is indeed csc x
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  8. #8
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    yeah but in this part, isnt the sin ^2 x is simplified to sin x so doesnt it have to be left in the numerator?

    Thanks!
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  9. #9
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    Quote Originally Posted by NeedHelp18 View Post


    yeah but in this part, isnt the sin ^2 x is simplified to sin x so doesnt it have to be left in the numerator?

    Thanks!
    If it was, sin(x) should be dividing the whole numerator...

    Here, it simplifies because $\displaystyle \cos^2 x+\sin^2 x=1$

    We don't want it to be more complicated
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  10. #10
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    thanks, now i get it
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