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Math Help - Simplifying trigonometric expressions

  1. #1
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    Simplifying trigonometric expressions

    I am having difficulties simplifying trigonometric expressions. My textbook isn't much help and I don't have a teacher to ask, so I thought maybe someone here could help me.

    Some examples of questions I'm not really sure about are:

    1) sinx / tanxcosx

    The answer I got for this one simply "1".

    2) sin(^4)x + (sin(^2)x)(cos(^2))

    This one I got "sinx(^4)x + 1"

    3)
    sinx - (sinx)(cos(^2)x)
    ------------------------
    sin(^2)x

    I am lost on this one.

    Any help would be greatly appreciated.
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  2. #2
    Junior Member
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    Here is number three...

    \frac{sin(x)-sin(x)*cos^2x}{sin^2x}

    First factor a sin(x) out of the numerator...

    \frac{sin(x)*(1-cos^2x)}{sin^2x}

    Then using the Pythagorean theorem we can say 1-cos^2x=sin^2x...

    \frac{sin(x)*sin^2(x)}{sin^2x}

    from there you can simplify to sin(x)
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  3. #3
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    Lexington, MA (USA)
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    Hello, KillamSin!

    1)\;\;\frac{\sin x}{\tan x\cos x}

    The answer I got for this one simply "1".
    You are right!

    \frac{\sin x}{\tan x\cos x} \;=\;\frac{\sin x}{\dfrac{\sin x}{{\color{red}\rlap{/////}}\cos x}({\color{red}\rlap{/////}}\cos x)} \;=\;\frac{{\color{green}\rlap{////}}\sin x}{{\color{green}\rlap{////}}\sin x} \;=\;1



    2)\;\; \sin^4\!x + \sin^2\!x\cos^2\!x

    \text{Factor: }\;\sin^2\!x\underbrace{\left(\sin^2\!x + \cos^2\!x\right)}_{\text{This is 1}} \;=\;\sin^2\!x




    3)\;\;\frac{\sin x - \sin x\cos^2\!x}{\sin^2\!x}

    \text{Factor: }\;\frac{\sin x\overbrace{\left(1 - \cos^2\!x\right)}^{\text{This is }\sin^2\!x}}{\sin^2\!x} \;=\;\frac{\sin x\cdot{\color{red}\rlap{/////}}\sin^2\!x}{{\color{red}\rlap{/////}}\sin^2\!x} \;=\;\sin x

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  4. #4
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    that is some perty LaTex! me+LaTex=
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