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Math Help - Trigonometric Identities and equations

  1. #1
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    Trigonometric Identities and equations

    Simplify each expression.

    5 c) 1-2sin^2 3π/8

    I got 1-2sin^2 67.5

    Cos2Θ= 1-2sin^2 67.5

    I dont know exactly what to do from there...
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Skoz View Post
    Simplify each expression.

    5 c) 1-2sin^2 3π/8

    I got 1-2sin^2 67.5

    Cos2Θ= 1-2sin^2 67.5

    I dont know exactly what to do from there...
    Leave it in terms of \pi otherwise you will never solve it

    1-2sin^2 \left(\frac{3\pi}{8}\right) = cos \left(\frac{3\pi}{4}\right) (because 2 \times \frac{3}{8} = \frac{3}{4})

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

    To solve use the angle difference formula: cos(A-B) = cosAcosB+sinAsinB

    Spoiler:
    Exact values of sin and cos at the two above values:

    cos(\pi) = -1
    sin(\pi) = 0

    cos \left(\frac{\pi}{4}\right) = sin \left(\frac{\pi}{4}\right) = \frac{\sqrt2}{2}
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  3. #3
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    where does the 3pi/4 come from?

    and how does cos (3pi/4)= (pi-pi/4) where did the 3 go?
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  4. #4
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    Quote Originally Posted by Skoz View Post




    where does the 3pi/4 come from?

    and how does cos (3pi/4)= (pi-pi/4) where did the 3 go?
    I used the double angle formula for cos:

    1-2sin^2 \theta = cos(2 \theta)

    Therefore 1-2sin^2 \left(\frac{3\pi}{8}\right) = cos \left(2 \cdot \frac{3\pi}{8}\right)

    \frac{3\pi}{4} comes from simplifying the cos term

    2 \times \frac{3\pi}{8} = \frac{6\pi}{8} = \frac{3\pi}{4}

    The 3 is turned into a 4-1. If you imagine \pi = \frac{4\pi}{4} then we see that
    \frac{4\pi}{4}-\frac{\pi}{4} = \frac{3\pi}{4}

    I used \pi and \frac{\pi}{4} because their exact values are known
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