Question 918988
Find sin2x, cos2x, and tan2x under the given conditions
cosx=-3/5, for pie < x < 3pie/2
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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
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Check:
cosx=-3/5(in quadrant III)
x=233.13&#730;
2x=466.26&#730;
..
sin2x=sin(466.26&#730;)&#8776;0.9600...
exact value=24/25=0.9600
..
cos2x=cos(466.26&#730;)&#8776;-0.2799...
exact value=-7/25=-0.2800
..
tan2x=tan(466.26&#730;)&#8776;-3.4286...
exact value=-24/7&#8776;-3.4285...