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Math Help - Simplifying Trigonometric Functions

  1. #1
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    Simplifying Trigonometric Functions

    I am having extreme troubles with this section. My book is no help and my teacher is clueless. Any help with these problems would be greatly helpful. Also if possible please include reasons/rules and show work to help me better understand and learn. Thanks!

    1). tan(Pi/2-x)cscx/csc^2x

    2). 1+tanx/1+cotx

    3). (sec^2x+csc^2x)-(tan^2x+cot^2x)

    4). sec^2u-tan^2u/cos^2v+sin^2v
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  2. #2
    Super Member bigwave's Avatar
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    hello bcavanaugh34
    welcome to the forum

    1). tan(Pi/2-x)cscx/csc^2x

    assuming this is

    <br />
\frac<br />
{\tan\left({\frac{\pi}{2}-x}\right)\csc{x}}<br />
{\csc^2{x}}<br />
\rightarrow<br />
\frac<br />
{\tan\left({\frac{\pi}{2}-x}\right)}<br />
{\csc{x}}<br />
\rightarrow<br />
\frac{\cot{x}}{\csc{x}}<br />
\rightarrow<br />
\frac{\cos{x}}{\sin{x}}\times\frac{\sin{x}}{1} = \cos{x}<br />

    hope this helps ..... still need more ?
    Last edited by bigwave; March 8th 2010 at 05:51 PM. Reason: latex
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  3. #3
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    More would be great. Thanks
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  4. #4
    Super Member bigwave's Avatar
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    #2

    \frac{1+\tan{x}}{1+cot{x}}<br />
\rightarrow<br />
\frac<br />
  {\frac <br />
       {\cos{x}+\sin{x}}{\cos{x}}}<br />
  {\frac<br />
{\sin{x}+\cos{x}}{\sin{x}}}<br />
\rightarrow<br />
{\frac <br />
       {\cos{x}+\sin{x}}{\cos{x}}}<br />
\times<br />
{\frac <br />
       {\sin{x}}{\cos{x}+\sin{x}}}<br />
\rightarrow<br />
\frac{\sin{x}}{\cos{x}}<br />
=\tan{x}<br /> <br />
    Last edited by bigwave; March 8th 2010 at 06:32 PM.
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  5. #5
    Super Member bigwave's Avatar
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    #3

    \left(\sec^2{\theta}+\csc^2{\theta}\right)<br />
-\left(\tan^2{\theta}+\cot^2{\theta}\right)<br />
\rightarrow<br />
\sec^2{\theta}+\csc^2{\theta}-\tan^2{\theta}-\cot^2{\theta}<br />
    note that
    \tan^2{\theta}+1=\sec^2{\theta}
    and
    1+\cot^2{\theta} = \csc^2{\theta}
    substituting these in:
    \tan^2{\theta}+1+1+\cot^2{\theta}-\tan^2{\theta}-\cot^2{\theta}=2<br />
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  6. #6
    Super Member bigwave's Avatar
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    #4

    sec^2u-tan^2u/cos^2v+sin^2v

    assuming this is

    \frac<br />
{\sec^2{u}-\tan^2{u}}<br />
{\cos^2{v}+\sin^2{v}}<br />
\rightarrow<br />
\frac{\sec^2{u}-(\sec^2{u}-1)}{1}=1<br /> <br />
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  7. #7
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    Thank you so much but i am having trouble with the last section. It says: use the basic identities to change the expression to one involving only sines and cosines. Then simplify to a basic trigometric function.

    1). (sinx)(tanx+cotx)

    2). sinx-tan(x)cos(x)+ cos(pi/2-x)

    3). sin(x)cos(x)tan(x)sec(x)csc(x)
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  8. #8
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    Hello, bcavanaugh34!

    The last two are almost too simple . . .


    3)\;\; \left(\sec^2\!x+\csc^2\!x\right)-\left(\tan^2\!x+\cot^2\!x\right)

    \text{We have: }\;\underbrace{\left(\sec^2\!x - \tan^2\!x\right)}_{\text{This is 1}} - \underbrace{\left(\csc^2\!x - \cot^2\!x\right)}_{\text{This is 1}} \;\;=\;\;1 + 1 \;\;=\;\;2



    4)\;\;\frac{\sec^2\!u-\tan^2\!u}{\cos^2\!v+\sin^2\!v}

    \text{We have: }\;\frac{\overbrace{\sec^2\!u - \tan^2\!u}^{\text{This is 1}}}{\underbrace{\cos^2\!v + \sin^2\!v}_{\text{This is 1}}} \;\;=\;\;\frac{1}{1} \;\;=\;\;1

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