SOLUTION: Find sin2x, cos2x, and tan2x under the given conditions cosx=-3/5, for 3pie/2 < x

SOLUTION: Find sin2x, cos2x, and tan2x under the given conditions cosx=-3/5, for 3pie/2 < x < 2pie

Question 918988: Find sin2x, cos2x, and tan2x under the given conditions
cosx=-3/5, for 3pie/2 < x < 2pie

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find sin2x, cos2x, and tan2x under the given conditions
cosx=-3/5, for pie < x < 3pie/2
***
You are working with a 3-4-5 reference right triangle in quadrant III where cos<0, sin<0
cosx=-3/5
sinx=-4/5
..
sin2x=2sinxcosx=2*-4/5*-3/5=24/25
cos2x=cos^2x-sin^2x=(-3/5)^2-(4/5)^2=9/25-16/25=-7/25
tan2x=sin2x/cos2x=-24/7
..
Check:
cosx=-3/5(in quadrant III)
x=233.13˚
2x=466.26˚
..
sin2x=sin(466.26˚)≈0.9600...
exact value=24/25=0.9600
..
cos2x=cos(466.26˚)≈-0.2799...
exact value=-7/25=-0.2800
..
tan2x=tan(466.26˚)≈-3.4286...
exact value=-24/7≈-3.4285...
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