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Math Help - Trigonometric function

  1. #1
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    Trigonometric function

    Given that tanx=p and sin2x=(2p)/(1+p^2). Without using table or calculator, find the value of tan15.

    I let tan15=p, tan30=(2p)/(1-p^2)
    1/(√3)= 2p/(1-p^2)
    1-p^2= 2
    √3p
    p^2 + 2
    √3p-1=0
    I still can't get the answer..
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  2. #2
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    Re: Trigonometric function

    Quote Originally Posted by Trefoil2727 View Post
    Given that tanx=p and sin2x=(2p)/(1+p^2). Without using table or calculator, find the value of tan15.
    I assume that x\in\text{quad }I then

    \sin(x)=\frac{p}{\sqrt{1+p^2}},~\cos(x)=\frac{1}{ \sqrt{1+p^2}},~\&~\sin(2x)=2\sin(x)\cos(x).
    Thanks from topsquark
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  3. #3
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    Re: Trigonometric function

    \displaystyle \begin{align*} \sin{(2x)} &= \frac{2p}{1 + p^2} \\ \\ \cos{(2x)} &= \sqrt{ 1 - \sin^2{(2x)}} \\ &= \sqrt{ 1 - \left( \frac{2p}{1 + p^2} \right) ^2 } \\ &= \sqrt{1 - \frac{4p^2}{ \left( 1 + p^2 \right) ^2 } } \\ &= \sqrt{ \frac{ \left(  1  + p^2 \right) ^2 - 4p^2 }{ \left( 1 + p^2 \right) ^2 } } \\ &= \sqrt{ \frac{1 + 2p^2 + p^4 - 4p^2}{ \left( 1 + p^2 \right) ^2 } } \\ &= \sqrt{ \frac{ 1 - 2p^2 + p^4 }{ \left( 1 + p^2 \right) ^2 } } \\ &= \sqrt{ \frac{ \left( 1 - p^2 \right) ^2 }{ \left( 1 + p^2 \right) ^2 } } \\ &= \frac{ 1 - p^2}{1 + p^2} \end{align*}

    \displaystyle \begin{align*} \tan{(2x)} &= \frac{\sin{(2x)}}{\cos{(2x)}} \\ &= \frac{\frac{2p}{1 + p^2}}{\frac{1 - p^2}{1 + p^2}} \\ &= \frac{2p}{1 - p^2}  \end{align*}

    Now if \displaystyle \begin{align*} x = 15^{\circ} \end{align*} we have \displaystyle \begin{align*} p = \tan{ \left( 15^{\circ} \right) } \end{align*}, so

    \displaystyle \begin{align*} \tan{ \left( 30^{\circ} \right) } &= \frac{2p}{1 - p^2} \\ \frac{1}{\sqrt{3}} &= \frac{2p}{1 - p^2} \\ 1 - p^2 &= 2\sqrt{3}\,p \\ p^2 + 2\sqrt{3}\,p - 1 &= 0 \\ p^2 + 2\sqrt{3}\,p + \left( \sqrt{3} \right) ^2 - \left( \sqrt{3} \right) ^2 - 1 &= 0 \\ \left( p + \sqrt{3} \right) ^2 - 4 &= 0 \\ \left( p + \sqrt{3} \right) ^2 &= 4 \\ p + \sqrt{3} &= \pm 2 \\ p &= -\sqrt{3} \pm 2   \end{align*}

    And since \displaystyle \begin{align*} 0^{\circ} < 15^{\circ} < 90^{\circ} \end{align*} that means \displaystyle \begin{align*} \tan{ \left( 15^{\circ} \right) } > 0 \end{align*}, which means the solution is \displaystyle \begin{align*} \tan{ \left( 15^{\circ} \right) } = 2 - \sqrt{3} \end{align*}.
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  4. #4
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    Re: Trigonometric function

    Tan x = p,
    sin 2x = 2p/(1+p^2) let x = 15, Then we have
    sin 30 = 2p /(1+p^2) = that gives
    p^2-4p+1=0
    Thus p = (4 √(16-12))/2 = (4 2√3)/2= 2 √3
    Because 0 < 15 < 90 that is in first quadrant, we have
    tan 15 = = 2- √3
    Thanks from Trefoil2727
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