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Math Help - PRECALCULUS (Trig Functions) -SO LOST

  1. #1
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    Post PRECALCULUS (Trig Functions) -SO LOST

    I am having a serous issue with these two problems. No matter what formula's I use i keep getting the wrong answer. I NEED HELP BADLY.

    1.) Use a sum or difference formula or a half angle formula to determine the value of the trigonometric functions. Give exact answers. Do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2);

    sin((11pi)/8) =
    sin(-(pi/12)) =
    cos((3pi)/8) =
    cos((11pi)/8) =

    &


    2.) If cos(t) = - (3/5) where pi < t < ((3pi)/2), find the values of the following trigonometric

    Give exact answers, do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as
    sqrt(2).

    cos(2t) =
    sin(2t) =
    cos(t/2) =
    sin(t/2) =

    Any and all Help greatly appreciated

    ~Kiba~
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Whitewolfblue View Post
    2.) If cos(t) = - (3/5) where pi < t < ((3pi)/2), find the values of the following trigonometric

    Give exact answers, do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as
    sqrt(2).

    cos(2t) =
    sin(2t) =
    cos(t/2) =
    sin(t/2) =
    cos(t) = -\frac{3}{5}

    so
    sin(t) = \pm \sqrt{1 - cos^2(t)} = \pm \sqrt{ \frac{16}{25} } = \pm \frac{4}{5}
    Since \pi < t < \frac{3 \pi}{2} we know to use the "-" sign. So
    sin(t) = - \frac{4}{5}

    Now.
    cos(2t) = 2cos^2(t) - 1 = 2 \frac{9}{25} - 1 = \frac{18 - 25}{25} = -\frac{7}{25}

    sin(2t) = 2 sin(t) cos(t) = 2 \cdot \frac{-4}{5} \cdot \frac{-3}{5} = \frac{24}{25}

    cos(t/2) = \pm \sqrt{ \frac{1 + cos(t)}{2} } = \pm \sqrt{ \frac{1 - \frac{3}{5}}{2} } = \pm \sqrt{\frac{1}{5}}

    Since \pi < t < \frac{3 \pi}{2} we know that \frac{ \pi}{2} < \frac{t}{2} < \frac{3 \pi}{4} < \pi so t/2 is in the second quadrant. This means we use the "-" sign.

    Thus
    cos(t/2) = - \frac{1}{\sqrt{5}}

    sin(t/2) = \pm \sqrt{ \frac{1 - cos(t)}{2} } = \pm \sqrt{ \frac{1 - -\frac{3}{5}}{2} } = \pm \sqrt{ \frac{4}{5} } = \pm \frac{2}{\sqrt{5}}

    Since t/2 is in the second quadrant, we use the "+" sign, so
    sin(t/2) = \frac{2}{\sqrt{5}}

    -Dan
    Last edited by topsquark; November 30th 2006 at 06:09 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Whitewolfblue View Post
    1.) Use a sum or difference formula or a half angle formula to determine the value of the trigonometric functions. Give exact answers. Do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2);

    sin((11pi)/8) =
    sin(-(pi/12)) =
    cos((3pi)/8) =
    cos((11pi)/8) =
    sin(11\pi/8) = sin(\pi + 3\pi/8) = sin(\pi)cos(3 \pi/8) + sin(3\pi/8)cos(\pi) = -sin(3\pi/8)

    = -sin \left ( \frac{1}{2} \cdot \frac{3 \pi}{4} \right )

    = - \sqrt{ \frac{1 - cos(3 \pi/4)}{2} } = - \sqrt{ \frac{1 + cos(\pi/4)}{2} } (Using the reference angle)

    = - \sqrt{ \frac{1 + \frac{\sqrt{2}}{2}}{2} } = - \sqrt{\frac{2 + \sqrt{2}}{4}}
    ================================================== =====

    cos(11\pi/8) = cos(\pi + 3\pi/8) = cos(\pi) cos(3\pi/8) - sin(\pi) sin(3\pi/8) = - cos(3\pi/8)

    = -cos \left ( \frac{1}{2} \cdot \frac{3 \pi}{4} \right )

    = - \sqrt{ \frac{1 + cos(3 \pi/4)}{2} } = -\sqrt{ \frac{1 - cos(\pi/4)}{2} }

    = - \sqrt{ \frac{1 - \frac{\sqrt{2}}{2}}{2} } = - \sqrt{ \frac{2 - \sqrt{2}}{4}}
    ================================================== =====

    sin(-\pi/12) = -sin \left ( \frac{1}{2} \cdot \frac{\pi}{6} \right )

    = - \sqrt{ \frac{1 - cos(\pi/6)}{2} } = -\sqrt{ \frac{1 - \frac{\sqrt{3}}{2}}{2} }

    = -\sqrt{\frac{2 - \sqrt{3}}{4} }

    And I already showed you what to do with cos(3 \pi/8) in the second problem.

    -Dan
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