Precalculus
posted by Jess .
If sinx= 1/4 and x terminates in the third quadrant, find the exact value of sin2x
My answer is sqrt15/8

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

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

I used that formula sin^2x+cos2x=1
1/4 squared minus that by one and squareroot it square root of 15 over 4 
you can do these a number of ways...

I get
cos(x)
=sqrt(1sin^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 
Awesome thanks mathmate!

You're welcome!