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Math Help - [SOLVED] Cosine of the Sum of Two Angles

  1. #1
    Newbie beardedoneder's Avatar
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    Exclamation [SOLVED] Cosine of the Sum of Two Angles

    Given sin \theta = \frac{-5}{13} in quadrant III, tan \phi = \frac{-8}{15} in quadrant II; find the value of sin( \theta + \phi):

    I've posted my solution -- which the answer key says is wrong -- here, http://imagebin.ca/img/tnINM0J.png .

    Please tell me what I'm doing wrong.
    Last edited by beardedoneder; April 7th 2010 at 04:19 PM. Reason: Post resolution, title "solved".
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  2. #2
    Super Member bigwave's Avatar
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    Cool

    Quote Originally Posted by beardedoneder View Post
    Given sin \theta = \frac{-5}{13} in quadrant III, tan \phi = \frac{-8}{15} in quadrant II; find the value of sin( \theta + \phi):

    I've posted my solution -- which the answer key says is wrong -- here, http://imagebin.ca/img/tnINM0J.png .

    Please tell me what I'm doing wrong.
    first if tan \phi = \frac{-8}{15}is in Q2
    or actually should be \tan\phi= \frac{8}{-15} if it is in Q2
    then \sin\phi = \frac{8}{17}

    therefore

    \sin(\theta + \phi)=\left(\text{-}\frac{5}{13}\right)\left(\text{-}\frac{15}{17}\right) + \left(\text{-}\frac{12}{13}\right)\left(\frac{8}{15}\right)<br />
\Rightarrow<br />
\frac{75}{221} - \frac{96}{221}<br />
\Rightarrow<br />
-\frac{21}{221}

    just got too late with latex
    Last edited by bigwave; March 17th 2010 at 07:33 PM. Reason: late with latex
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  3. #3
    MHF Contributor
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    Quote Originally Posted by beardedoneder View Post
    Given sin \theta = \frac{-5}{13} in quadrant III, tan \phi = \frac{-8}{15} in quadrant II; find the value of sin( \theta + \phi):

    I've posted my solution -- which the answer key says is wrong -- here, http://imagebin.ca/img/tnINM0J.png .

    Please tell me what I'm doing wrong.
    I agree w/ your solution.
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  4. #4
    Master Of Puppets
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    I would do it like this

    Quote Originally Posted by beardedoneder View Post
    Given sin \theta = \frac{-5}{13} in quadrant III,
    use \sin^2\theta+\cos^2\theta = 1 to find \cos\theta


    Quote Originally Posted by beardedoneder View Post
    tan \phi = \frac{-8}{15} in quadrant II;
    use 1+\tan^2\phi = \sec^2\phi to find \cos\phi

    and then \sin^2\phi+\cos^2\phi = 1 to find \sin\phi
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  5. #5
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    Hello, beardedoneder!

    Sorry, I couldn't read your work.


    Given: . \begin{array}{ccc}\sin\theta \:=\:\text{-}\frac{5}{13}& \text{in quadrant III} \\ \\ [-3mm]<br />
\tan\phi \:=\:\text{-}\frac{8}{15} & \text{in quadrant II}\end{array}

    find the value of: . \sin(\theta + \phi)

    We have: . \sin\theta \:=\:-\frac{5}{13} \:=\:\frac{opp}{hyp}

    \theta is in a right triangle with: opp = -5,\;hyp = 13

    Pythagorus says: . adj = \pm12

    . . In Quadrant III: adj = -12

    Hence: . \begin{Bmatrix} \sin\theta &=& \text{-}\dfrac{5}{13} \\ \\[-3mm]  \cos\theta &=& \text{-}\dfrac{12}{13} \end{Bmatrix} .[1]


    We have: . \tan \phi \:=\:-\frac{8}{15} \:=\:\frac{opp}{adj}

    \phi is in Quadrant II with: opp =  8,\;adj = -15

    Pythagorus says: . hyp = 17

    Hence: . \begin{Bmatrix}\sin\phi &=& \dfrac{8}{17} \\ \\[-3mm] \cos\phi &=& \text{-}\dfrac{15}{17} \end{Bmatrix} .[2]



    We want: . \sin(\theta + \phi) \;=\;\sin\theta\cos\phi + \cos\theta\sin\phi


    Substitute [1] and [2]:

    \sin(\theta + \phi) \;=\;\left(\text{-}\frac{5}{13}\right)\left(\text{-}\frac{15}{17}\right) + \left(\text{-}\frac{12}{13}\right)\left(\frac{8}{15}\right) \;=\;\frac{75}{221} - \frac{96}{221} \;=\;-\frac{21}{221}

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  6. #6
    Newbie beardedoneder's Avatar
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    Answer Key Was Wrong

    Quote Originally Posted by beardedoneder View Post
    Given sin \theta = \frac{-5}{13} in quadrant III, tan \phi = \frac{-8}{15} in quadrant II; find the value of sin( \theta + \phi):

    I've posted my solution -- which the answer key says is wrong -- here, http://imagebin.ca/img/tnINM0J.png .

    Please tell me what I'm doing wrong.
    I got a copy of the Solution Key. It solves the problem the same way I do, but it gives a product of \frac{225}{221} for -\frac{5}{13} \cdot -\frac{15}{17} instead of \frac{75}{221}. Which is why the answer key gives an answer of \frac{129}{221}, instead of -\frac{21}{221}. My solution and answer were correct.
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  7. #7
    Newbie beardedoneder's Avatar
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    Did you mean to say, ...

    Quote Originally Posted by bigwave View Post
    first if tan \phi = \frac{-8}{15}is in Q2
    or actually should be \tan\phi= \frac{8}{-15} if it is in Q2
    then \sin\phi = \frac{8}{17}

    therefore

    \sin(\theta + \phi)=\left(\text{-}\frac{5}{13}\right)\left(\text{-}\frac{15}{17}\right) + \left(\text{-}\frac{12}{13}\right)\left(\frac{8}{15}\right)<br />
\Rightarrow<br />
\frac{75}{221} - \frac{96}{221}<br />
\Rightarrow<br />
-\frac{21}{221}

    just got too late with latex

    Did you mean to say, -\frac{12}{13} \cdot \frac{8}{17} ?
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  8. #8
    Newbie beardedoneder's Avatar
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    Did you mean to say, ...

    Quote Originally Posted by Soroban View Post
    Hello, beardedoneder!

    Substitute [1] and [2]:

    \sin(\theta + \phi) \;=\;\left(\text{-}\frac{5}{13}\right)\left(\text{-}\frac{15}{17}\right) + \left(\text{-}\frac{12}{13}\right)\left(\frac{8}{15}\right) \;=\;\frac{75}{221} - \frac{96}{221} \;=\;-\frac{21}{221}


    Did you mean to say, -\frac{12}{13} \cdot \frac{8}{17} ?
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