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December 21, 2014

December 21, 2014

Posted by **Jess** on Saturday, August 6, 2011 at 6:14pm.

My answer is sqrt-15/8

- Precalculus -
**bobpursley**, Saturday, August 6, 2011 at 6:25pmsin2x=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 -
**Jess**, Saturday, August 6, 2011 at 6:31pmwait i dont understand how you did your math?

- Precalculus -
**Jess**, Saturday, August 6, 2011 at 6:34pmdont you use the formula sin2 theta =2sintheta cos theta?

- Precalculus -
**Jess**, Saturday, August 6, 2011 at 6:36pmI used that formula sin^2x+cos2x=1

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

- Precalculus -
**bobpursley**, Saturday, August 6, 2011 at 6:55pmyou can do these a number of ways...

- Precalculus -
**MathMate**, Saturday, August 6, 2011 at 7:16pmI 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 -
**Jess**, Saturday, August 6, 2011 at 7:29pmAwesome thanks mathmate!

- Precalculus :) -
**MathMate**, Saturday, August 6, 2011 at 7:42pmYou're welcome!

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