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Math Help - Trig Identities Problem

  1. #1
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    Trig Identities Problem

    cos(arcTan (5/12) - arcCos(2/3))
    so basically I set up the two triangles and get the hypotenuse 13 and the side length 2sqrt5. I'm supposed to evaluate without using a calculator, what do I do next?
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  2. #2
    MHF Contributor ebaines's Avatar
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    First - recognize that what you have are two angles - call them alpha and beta, where:

    <br />
\alpha = tan ^{-1} (\frac 5 {12}), \ \ \beta = cos ^{-1} (\frac 2 3 )<br />

    You are aked to find  cos(\alpha - \beta) . So use the trig identity:  cos(\alpha - \beta) = cos \alpha \cdot \cos \beta + sin \alpha \cdot sin \beta

    You can figure the values of these sines and cosines from the information you've been given. For example,  cos \alpha = \frac {12} {13} . BTW - the side legth of the second triangle is  \sqrt { 3^2 - 2^2} = \sqrt 5
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  3. #3
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    Hello, hellojellojw!

    Smplify: . \cos\left(\arctan\tfrac{5}{12} - \arccos\tfrac{2}{3}\right) .[1]

    Let: . \alpha \:=\: \arctan\tfrac{5}{12} \quad\Rightarrow\quad \tan\alpha \:=\:\tfrac{5}{12} \:=\:\tfrac{opp}{hyp}

    We have: . opp = 5,\;adj = 12 \quad\Rightarrow\quad hyp = 13

    . . Hence: . \sin\alpha \,=\,\tfrac{5}{13},\;\cos\alpha \,=\,\tfrac{12}{13} .[2]


    Let: . \beta \,=\,\arccos\tfrac{2}{3} \quad\Rightarrow\quad \cos\beta \:=\:\tfrac{2}{3} \:=\:\tfrac{adj}{hyp}

    We have: . adj = 2,\;hyp = 3 \quad\Rightarrow\quad opp = \sqrt{5}

    . . Hence: . \sin\beta \,=\,\tfrac{\sqrt{5}}{3},\;\cos\beta \,=\,\tfrac{2}{3} .[3]


    From [1], we have: . \cos(\alpha - \beta) \;=\;\cos\alpha\cos\beta + \sin\alpha\sin\beta


    Substitute [2] and [3]: . \cos(\alpha + \beta) \;=\;\left(\tfrac{12}{13}\right)\left(\tfrac{2}{3}  \right) + \left(\tfrac{5}{13}\right)\left(\tfrac{\sqrt{5}}{3  }\right) \;=\;\frac{24+5\sqrt{5}}{39}

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  4. #4
    Junior Member BugzLooney's Avatar
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    why does arctan being 5/12 and arccos being 2/3 imply that the tangent and cosine are also those values?
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  5. #5
    MHF Contributor ebaines's Avatar
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    Quote Originally Posted by BugzLooney View Post
    why does arctan being 5/12 and arccos being 2/3 imply that the tangent and cosine are also those values?
    Definition of the arctan: it's the angle whose tangent is the given value. Think of arctan as the inverse function of tan. Hence the tangent of the arctangent of x is simply x. You can see this with a simple example: consider the angle whose tangent = 1. You can probably see right away that the angle is  \pi/4 . Now find the tangent of  \pi/4 - it's 1. So you see that tan(arctan(x)) = x.
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