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Precalculus

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If sinx= -1/4 and x terminates in the third quadrant, find the exact value of sin2x


My answer is sqrt-15/8

  • Precalculus - ,

    sin2x=2sinx cos x

    = 2*-.25*cos x but

    sin^2x+cos^2x=1 so
    cos^2x=.75
    cosx=sqrt 3/4

    so now you have it.
    check sign: if x is in the third quadrant, so 2x will be in the second quadrant, where sine is +

  • Precalculus - ,

    wait i don't understand how you did your math?

  • Precalculus - ,

    dont you use the formula sin2 theta =2sintheta cos theta?

  • Precalculus - ,

    I used that formula sin^2x+cos2x=1

    -1/4 squared minus that by one and squareroot it square root of 15 over 4

  • Precalculus - ,

    you can do these a number of ways...

  • Precalculus - ,

    I get
    cos(x)
    =sqrt(1-sin^2(x))
    =sqrt(1-(1/4)^2)
    =sqrt(15/16)

    so
    sin(2x)
    =+2sin(x)cos(x)
    =+2*(1/4)sqrt(15/16)
    =sqrt(15) / 8

  • Precalculus - ,

    Awesome thanks mathmate!

  • Precalculus :) - ,

    You're welcome!

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