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Math Help - trigonometry; solve

  1. #1
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    trigonometry; solve

    if tan θ = 2/5 for 0< θ <360^2, find the values of sin θ
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  2. #2
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    Quote Originally Posted by deej813 View Post
    if tan θ = 2/5 for 0< θ <360^2, find the values of sin θ
    1. \tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac25

    2. \sin^2(\theta) + \cos^2(\theta)=1

    3. Substitute:

    \tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{\sin(\theta)}{\sqrt{1-\sin^2(\theta)}} = \dfrac25

    4. Solve for \sin(\theta) .
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  3. #3
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    should i get to
    25sin^2θ=1-sinθ
    where do i go from there??
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  4. #4
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    Quote Originally Posted by deej813 View Post
    hey ive really stuffed this up
    can anyone show me what to do <<<< the only adequate answer to this question is: Yes!
    Obviously solving equations isn't very familiar to you. So here we go:

     \dfrac{\sin(\theta)}{\sqrt{1-\sin^2(\theta)}} = \dfrac25 Multiply both sides of the equation by the complete square-root:

     \sin(\theta) = \dfrac25 \cdot \sqrt{1-\sin^2(\theta)} Square both sides:

     \sin^2(\theta) = \dfrac4{25} \cdot (1-\sin^2(\theta)) Expand the bracket:

     \sin^2(\theta) = \dfrac4{25}- \dfrac4{25} \cdot \sin^2(\theta) Collect like terms on one side:

    \dfrac{29}{25} \cdot \sin^2(\theta) = \dfrac4{25} Divide both sides by \dfrac{29}{25} :

    \sin^2(\theta) = \dfrac4{29} Calculate the square root at both sides:

    | \sin(\theta) | = \sqrt{\dfrac4{29}} = \dfrac2{29} \cdot \sqrt{29}

    Due to the squaring in step #2 you'll get 2 solutions of this equation:

     \sin(\theta) = \dfrac2{29} \cdot \sqrt{29}~\vee~ \sin(\theta) =  - \dfrac2{29} \cdot \sqrt{29}

    You must check if one or both are valid.
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  5. #5
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    tangent function is positive in quadrants I and III. So, the hypotenuse is equal to square root of (5^2 + 2^2) = square root of 29. Then sin theta is equal to + 2/(sqrt of 29) and - 2/(sqrt of 29), because for sin theta, positive in quadrant I and negative in quadrant III.
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  6. #6
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    Hello, deej813!

    if \tan \theta = \tfrac{2}{5}\;\text{ for }0< \theta  <360^o, find the values of \sin \theta
    Since \tan\theta is positive, \theta is in Quadrant 1 or 3.


    We have: . \tan\theta \:=\:\frac{2}{5} \:=\:\frac{opp}{adj}

    Assume \theta is an acute angle.
    Then \theta is in a right triangle with: . opp = 2,\;adj = 5
    Pythagorus says: . hyp \:=\:\sqrt{2^2 + 5^2} \:=\:\sqrt{29}


    In Quadrant 1, we have: . \begin{Bmatrix}opp &=& 2 \\ adj &=& 5 \\ hyp &=& \sqrt{29}\end{Bmatrix}

    . . Hence: . \sin\theta \:=\:\frac{opp}{adj} \;=\;\frac{2}{\sqrt{29}}


    In Quadrant 3, we have: . \begin{Bmatrix}opp &=& \text{-}2 \\ adj &=& \text{-}5 \\ hyp &=& \sqrt{29} \end{Bmatrix}

    . . Hence: . \sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{-2}{\sqrt{29}}


    Therefore: . \boxed{\sin\theta \;=\;\pm\frac{2}{\sqrt{29}}}

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