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Math Help - Trigonometric Identities

  1. #1
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    Trigonometric Identities

    plz answer dis question......


    Find the value of sin 15o = sin ( π12 )

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  2. #2
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    Re: Trigonometric Identities

    You should know that \displaystyle \begin{align*} \sin{(2\theta)} = 2\sin{(\theta)}\cos{(\theta)} = 2\sin{(\theta)}\,\sqrt{1 - \sin^2{(\theta)} } \end{align*}. So

    \displaystyle \begin{align*} \sin{ \left( 30^{\circ} \right) } &= 2\sin{ \left( 15^{\circ} \right) } \, \sqrt{ 1 - \sin^2{ \left( 15^{\circ} \right) } } \\ \frac{1}{2} &= 2\sin{ \left( 15^{\circ} \right) } \, \sqrt{ 1 - \sin^2{ \left( 15^{\circ} \right) } } \\ \frac{1}{4} &= \sin{ \left( 15^{\circ} \right) } \, \sqrt{ 1 - \sin^2{ \left( 15^{\circ} \right) } } \\ \left( \frac{1}{4} \right) ^2 &= \left[ \sin{ \left( 15^{\circ} \right) } \, \sqrt{ 1 - \sin^2{ \left( 15^{\circ} \right) } } \right] ^2 \\ \frac{1}{16} &= \sin^2{ \left( 15^{\circ} \right) } \left[ 1 - \sin^2{ \left( 15^{\circ} \right) } \right] \\ \frac{1}{16} &= \sin^2{ \left( 15^{\circ} \right) } - \sin^4{ \left( 15^{\circ} \right) } \\ \sin^4{ \left( 15^{\circ} \right) } - \sin^2{ \left( 15^{\circ} \right) } + \frac{1}{16} &= 0 \\ x^2 - x + \frac{1}{16} &= 0 \textrm{ if we let } x = \sin^2{ \left( 15^{\circ} \right) } \\ x &= \frac{1 \pm \sqrt{ (-1)^2 - 4(1)\left( \frac{1}{16} \right) } }{2(1)} \end{align*}

    \displaystyle \begin{align*} x &= \frac{1 \pm \sqrt{ 1 - \frac{1}{4}}}{2} \\ x &= \frac{1 \pm \sqrt{ \frac{3}{4} } }{2} \\ x &= \frac{1 \pm \frac{\sqrt{3}}{2}}{2} \\ x &= \frac{ \frac{2 \pm \sqrt{3}}{2} }{2} \\ x &= \frac{2 \pm \sqrt{3}}{4} \\ \sin^2{ \left( 15^{\circ} \right) } &= \frac{2 \pm \sqrt{3}}{4} \\ \sin{ \left( 15^{\circ} \right) } &= \pm \frac{\sqrt{2 \pm \sqrt{3}}}{2} \end{align*}

    When we note that sine is positive in the first quadrant, and check for extraneous solutions which were obtained through squaring, we find that the solution we want is \displaystyle \begin{align*} \sin{ \left( 15^{\circ} \right) } = \frac{\sqrt{2 - \sqrt{3}}}{2} \end{align*}.
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    Re: Trigonometric Identities

    sin 15° = sin(45°-30°)
    = sin 45°cos 30° - cos 45°sin 30°
    = \[(1/\sqrt 2 *\sqrt 3 /2) - (1/2*1/\sqrt {2)}  = (\sqrt 6  - \sqrt 2 )/4\]
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    Re: Trigonometric Identities

    Quote Originally Posted by mpx86 View Post
    sin 15° = sin(45°-30°)
    = sin 45°cos 30° - cos 45°sin 30°
    = \[(1/\sqrt 2 *\sqrt 3 /2) - (1/2*1/\sqrt {2)}  = (\sqrt 6  - \sqrt 2 )/4\]
    How DARE you provide such a simple, concise and beautiful solution after mine took 18 lines? :P hahaha. Well played
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