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Math Help - [SOLVED] Trig equation - Why can't I get it right?!?

  1. #1
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    Exclamation [SOLVED] Trig equation - Why can't I get it right?!?

    Hi. I've been trying to solve this trig. equation 4sin4t + 3cos4t = 0. The answer is meant to be t = \frac{\pi}{4}-\frac{1}{4}\arctan(\frac{3}{4})

     <br />
4sin4t + 3cos4t = 0<br />

     <br />
4sin4t = - 3cos4t<br />

     <br />
tan4t = -\frac{3}{4}<br />

     <br />
4t = \arctan(\frac{3}{4})<br />

     <br />
t = \frac{1}{4}\arctan(\frac{3}{4})<br />
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by UltraGirl View Post
    Hi. I've been trying to solve this trig. equation 4sin4t + 3cos4t = 0. The answer is meant to be t = \frac{\pi}{4}-\frac{1}{4}\arctan(\frac{3}{4})

     <br />
4sin4t + 3cos4t = 0<br />

     <br />
4sin4t = - 3cos4t<br />

     <br />
tan4t = -\frac{3}{4}<br />

     <br />
4t = \arctan({\color{red}-}\frac{3}{4})={\color{red}-}\arctan\frac{3}{4}<br />
    You have dropped the minus-sign, also note that there are infinitely many possibilities, namely

    4t=-\arctan\frac{3}{4}+n\cdot\pi, n\in\mathbb{Z}

    So by dividing by 4 you get the general solution

    t =n\cdot \frac{\pi}{4}-\arctan\frac{3}{4}

    Now plug in 1 for n and you get the particular solution that you have been given.
    Last edited by Failure; May 5th 2010 at 03:05 AM.
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  3. #3
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    Quote Originally Posted by UltraGirl View Post
    Hi. I've been trying to solve this trig. equation 4sin4t + 3cos4t = 0. The answer is meant to be t = \frac{\pi}{4}-\frac{1}{4}\arctan(\frac{3}{4})

     <br />
4sin4t + 3cos4t = 0<br />

     <br />
4sin4t = - 3cos4t<br />

     <br />
tan4t = -\frac{3}{4}  <br />
specifies a line through the origin with a slope of -3/4

     <br />
4t = \arctan(\frac{3}{4}) <br /> <br />
gives you the "acute" angle the line makes with the negative part of the x-axis in the 2nd quadrant

    You need to specify the counterclockwise angle the line makes with the positive half of the x-axis

     <br />
t = \frac{1}{4}\arctan(\frac{3}{4})
    Hi Ultragirl,

    4sin4t+3cos4t=0

    tan4t=-\frac{3}{4}

    tan4t=tan^{-1}\left(-\frac{3}{4}\right)

    Line through origin, with slope -\frac{3}{4} passes through 2nd and 4th quadrants.

    The angle tan^{-1}\left(\frac{3}{4}\right) is the acute angle the line makes with the negative x-axis in the 2nd quadrant
    or with the positive x-axis in the 4th quadrant.

    The angle 4t is the pair of counterclockwise angles the line with negative slope
    makes with the positive x-axis.

    In the 2nd quadrant...

    4t={\pi}-tan^{-1}\left(\frac{3}{4}\right)

    t=\frac{{\pi}}{4}-\frac{1}{4}tan^{-1}\left(\frac{3}{4}\right)

    For other solutions, add integer multiples of {\pi}

    In the attachment,

    4t\ =\ counterclockwise\ angles\ AOB\ and\ AOC

    \theta=arctan\left(\frac{3}{4}\right)
    Attached Thumbnails Attached Thumbnails [SOLVED] Trig equation - Why can't I get it right?!?-inverse-tan.jpg  
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