How do you find the value of #sin(theta/2)# given #costheta=-4/5# and #90<theta<180#?

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Answer 1 · 2017-01-26T13:49:50

#sin(theta/2)=+3/sqrt10.#

Recall that #cos2x=1-2sin^2x.#

#:. cos theta=1-2sin^2(theta/2)#

#:. 2sin^2(theta/2)=1-(-4/5)=9/5#

#:. sin^2(theta/2)=9/10#

#:. sin(theta/2)=+-3/sqrt10#

But, #90 lt theta lt 180 rArr 90/2 lt theta/2 lt 180/2, i.e.,#

#45 lt theta/2 lt 90," or, "theta/2" lies in the "1^(st)" Quadrant".#

#"Therefore, "sin(theta/2)=+3/sqrt10.#