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Math Help - Compound Angle Formulas - Confused over evaluating

  1. #1
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    Compound Angle Formulas - Confused over evaluating

    I have done questions like these before, but the ones I did were only in quadrant I. I really wish I knew how to draw the triangles on here for you, so you see everything that I am working with.

    Question:
    If x is in the interval  (\frac{\pi}{2}, \pi) and y is in the interval  (\pi, \frac {3\pi}{2}) and cosx =  - \frac {5}{13} and tany =  \frac {4} {3} , evaluate the following.

    *I will put the thing to evaluate, then the answer (according to the sheet) under it, then put my attempt after*

    a) sin (x + y)
    Answer:  - \frac{16}{65}

    b) cos (x - y)
    Answer:  - \frac {33}{65}

    c) tan (x - y)
    Answer:  \frac {56}{33}



    Okay, now I will show my attempts (for some reason I got the answer for b, as the sheet answer for a. And the answer for b as the sheet answer for a.)


    a)  \sin (x + y)
    =  \sin x \cos y + \cos x \sin y
    =  ((\frac {12}{13}) (\frac {-4}{-5})) + ((\frac {-5}{13}) (\frac {-3}{-5})
    =  (\frac {48}{65}) + (\frac{-3}{13})
    =  \frac {33}{65}


    b)  cos (x - y)
    =  cosx cosy + sinx siny
    =  ((\frac{-5}{13})(\frac{-4}{5}) + ((\frac {12}{13})(\frac{-3}{5}))
    =  \frac {4}{13} + \frac {-36}{65}
    =  \frac {-16}{65}

    c)  tan (x - y)
    =  \frac {tanx - tany} {1 + tanxtany}
    =  \frac {(\frac {12}{13}) - \frac {-3}{-4}} {1 + (\frac {12}{13}) (\frac {-3}{-4}) }
    =  \frac { \frac {9}{52} } {1+ \frac {9}{13} \frac {22}{13}}
    =  \frac {9} {88}
    Last edited by Xenophobe; April 19th 2009 at 03:16 PM.
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  2. #2
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    Hello, Xenophobe!

    If x is in the interval \left(\tfrac{\pi}{2}, \pi\right) and y is in the interval \left(\pi, \tfrac{3\pi}{2}\right)

    and \cos x = -\tfrac{5}{13} and \tan y = \tfrac{4}{3} , evaluate the following:

    (a)\;\sin(x + y) \qquad (b)\;\cos(x - y) \qquad (c)\;\tan (x - y)

    x is in Quadrant 2, and \cos x \:=\: -\frac{5}{13} \:=\:\frac{adj}{hyp}
    We have: . adj = -5,\;hyp = 13
    . . Using Pythagorus: . opp \:=\:12


    y is in Quadrant 3, and \tan y \:=\:\frac{4}{3} \:=\:\frac{opp}{adj}

    In Quadrant 3: . opp = -4,\; adj = -3,\; hyp = 5


    We have these values: . \begin{array}{cccccccc}<br />
\sin x &=& \dfrac{12}{13} & & \sin y &=& \text{-}\dfrac{4}{5} \\ \\[-3mm]<br />
\cos x &=& \text{-}\dfrac{5}{13} & & \cos y &=& \text{-}\dfrac{3}{5} \\ \\[-3mm]<br />
\tan x &=& \text{-}\dfrac{12}{5} & & \tan y &=& \dfrac{4}{3} \end{array}


    Try those in your compound angle formulas . . .

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