Given $costheta=-4/5$ and $-270<theta<-180$ , how do you find $sin (theta/2)$ ?

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Answer 1 · 2016-06-23T13:01:39

$sin(theta/2)=3/sqrt10$

Let $A=theta/2$ , as $cos2A=1-2sin^2A$ , we have

$costheta=1-2sin^2(theta/2)$

or $-4/5=1-2sin^2(theta/2)$

or $2sin^2(theta/2)=1+4/5=9/5$

or $sin^2(theta/2)=9/10$

As $-270^o < theta < -180^o$ , $theta$ is in second quadrant and is positive

Hence $sin(theta/2)=sqrt(9/10)=3/sqrt10$

Given costheta=-4/5costheta=-4/5 and -270<theta<-180-270<theta<-180, how do you find sin (theta/2)sin (theta/2)?