Given #sintheta=-9/22# and #270<theta<360#, how do you find #sin2theta#?

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Answer 1 · 2016-07-08T12:49:00

#-9sqrt(403)/242#

you know that

#sin2theta=2sinthetacostheta#

since #sintheta=-9/22#

and you know that, in the fourth quadrant:

#costheta=sqrt(1-sin^2theta)#

and, in this case,

#costheta=sqrt(1-(-9/22)^2)#

#=sqrt(1-81/484)#

#=sqrt(403)/22#

so

#sin2theta=2(-9/22)(sqrt(403)/22)#

#=-9sqrt(403)/242#