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Math Help - Quick sin,cos,tan question

  1. #1
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    Quick sin,cos,tan question

    Given tan(theta) = 8, find cos(theta) and sin(theta)

    I have no idea how to do this lol, i forgot >_< and i want exact answers plz >_< so no decimals 0_o
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  2. #2
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    think of the right triangle.
    \cos\theta=\frac{adjacent}{hypotenuse}
    \sin\theta=\frac{opposite}{hypotenuse}
    \tan\theta=\frac{opposite}{adjacent}
    Also Pythagorus' theorem: a^2=b^2+c^2
    where a is the hypotenuse, and b and c are adjacent and opposite sides respectively.
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  3. #3
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    Essentially the same: think of tan(\theta)= 8 as representing a right triangle with "near side" of length 8 and "opposite side" of length 1 so that tan(\theta)= "near side"/"opposite side"= 8. Use the Pythagorean theorem to find the hypotenuse and then use the formulas arze gives.
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  4. #4
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    Quote Originally Posted by haddad287 View Post
    Given tan(theta) = 8, find cos(theta) and sin(theta)

    I have no idea how to do this lol, i forgot >_< and i want exact answers plz >_< so no decimals 0_o
    You haven't defined your domain, so you may also consider Tan\theta to be the slope of a line through the origin, for 0\ \le\ \theta\ \le\ 2{\pi}.

    Hence, there are also negative solutions for Sin\theta and Cos\theta.


    You don't know the lengths of the sides of your right-angled triangle, but all you need is the ratios.

    Adjacent=x

    Opposite=8x

    Hypotenuse=\sqrt{(8x)^2+x^2}=x\sqrt{65}
    Last edited by Archie Meade; September 16th 2010 at 07:01 AM.
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  5. #5
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    Hello, haddad287!

    \text{Given: }\:\tan\theta \,=\, 8

    \text{Find: }\;\cos\theta\,\text{ and }\,\sin\theta

    I will assume that \,\theta is an acute angle.


    \text{We have: }\;\tan\theta \:=\:\dfrac{8}{1} \:=\:\dfrac{opp}{adj}

    That is: . \,\theta is in a right triangle with opp = 8,\;adj = 1
    . .
    (Can you picture that? If not, make a sketch.)

    Pythagorus tell us that: . hyp \:=\:\sqrt{8^2+1^2} \:=\:\sqrt{65}


    Therefore: . \begin{Bmatrix}\cos\theta &=& \dfrac{adj}{hyp} &=& \dfrac{1}{\sqrt{65}} \\ \\[-3mm] \sin\theta &=& \dfrac{opp}{hyp} &=& \dfrac{8}{\sqrt{65}} \end{Bmatrix}
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  6. #6
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    Quote Originally Posted by haddad287 View Post
    Given tan(theta) = 8, find cos(theta) and sin(theta)

    I have no idea how to do this lol, i forgot >_< and i want exact answers plz >_< so no decimals 0_o
    Here is a different approach.
    <br />
\tan\theta= \frac{\sin\theta}{\cos\theta} =8

    \tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=64

    \sin^2\theta=64 \cos^2\theta

    \sin^2\theta =64 (1-\sin^2\theta)

    \sin^2\theta =64-64\sin^2\theta

    65\sin^2\theta = 64

    \sin^2\theta = \frac{64}{65}

    \sin\theta = \frac{8}{\sqrt{65}}


    <br />
\tan\theta= \frac{\sin\theta}{\cos\theta} =8

    \tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=64

    \frac{1-\cos^2\theta}{\cos^2\theta}=64
    <br />
\frac{1}{\cos^2\theta}-\frac{\cos^2\theta}{\cos^2\theta}=64

    \frac{1}{\cos^2\theta}-1=64

    \frac{1}{\cos^2\theta}=65

    \cos^2\theta=\frac{1}{65}

    \cos\theta=\frac{1}{\sqrt{65}}
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