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Math Help - Trigonometry Identities? Please Help.

  1. #1
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    Trigonometry Identities? Please Help.

    How do i solve this equation?

    sin(2x) = -(sqrt(2))/2

    x = ___ ?

    ------------------

    i think i have to use: sin(2x) = 2sinxcosx
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  2. #2
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    \displaystyle x=\frac{sin^{-1}\left(\frac{-\sqrt{2}}{2}\right)}{2}+2\pi k \ \k\in\mathbb{Z}
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  3. #3
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    Hello, vecoma!

    How do i solve this equation?

    . . \sin(2x) \:=\: -\dfrac{\sqrt{2}}{2}

    \text{We have: }\:\sin(2x) \:=\:-\dfrac{1}{\sqrt{2}}


    From your basic trig values, you should know that:

    . . the angle whose sine is \text{-}\frac{1}{\sqrt{2}}\,\text{ is }\,\dfrac{5\pi}{4}\,\text{ or }\,\dfrac{7\pi}{4}


    \text{So we have: }\;2x \;=\;\begin{Bmatrix}\dfrac{5\pi}{4} + 2\pi n \\ \\[-3mm] \dfrac{7\pi}{4} + 2\pi n \end{Bmatrix}


    . . \text{Therefore: }\;x \;=\;\begin{Bmatrix}\dfrac{5\pi}{8} + \pi n \\ \\[-3mm] \dfrac{7\pi}{8} + \pi n \end{Bmatrix}\;\text{ for some integer }n

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  4. #4
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    Quote Originally Posted by vecoma View Post
    How do i solve this equation?

    sin(2x) = -(sqrt(2))/2

    x = ___ ?

    ------------------

    i think i have to use: sin(2x) = 2sinxcosx
    Hi vecoma,

    sin2x=2sinxcosx expresses the sine of a double angle in terms of the single angle.

    It only introduces complexity to your problem as you now have 2 trigonometric terms instead of 1.

    Instead sinA=-\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{\sqrt{2}\;\sqrt{2}}

    where A=2x

    first find A, then x
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