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Thread: Trigonometric Identities

  1. September 13th 2008, 11:18 AM #1
    robasc
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    Trigonometric Identities

    I need help trying to determine what number is suppose to be in place for alpha? I do not understand how they came up with Cos = 1/4 in question (d).

    Sin(90 deg - alpha) = Cos alpha = 1/4

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  2. September 13th 2008, 11:32 AM #2
    Moo
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    Hello,

    \sin(90°-\alpha)=\cos(\alpha) can be checked on a unit circle.

    Otherwise, use this formula : \sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)

    So here \sin(90°-\alpha)={\color{red}\sin(90°)}\cos(\alpha)-\sin(\alpha) {\color{red}\cos(90°)}

    The red parts are supposed to be known
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