How do you find the value of #sin(theta/2)# given #costheta=(2sqrt5)/5# and #0<theta<90#?

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Answer 1 · 2016-07-20T18:42:32

#sin(theta/2)=sqrt{(5-2sqrt5)/10}~=0.2297#

We use the Identity # : costheta=1-2sin^2(theta/2)#

#:. 2sqrt5/5=1-2sin^2(theta/2)#

#:. 2sin^2(theta/2)=1-2sqrt5/5=(5-2sqrt5)/5#

#:. sin^2(theta/2)=(5-2sqrt5)/10#

Now, #(5-2sqrt5)/10~=(5-2(2.2361))/10#

#=(5-4.4722)/10=0.5278/10=0.05278#

Hence, #sin(theta/2)=sqrt{(5-2sqrt5)/10}~=0.2297#

While taking square-root, the #-ve# sign was deliberately avoided as

#0<,theta<,90 rArr 0<,theta/2<,45rArr sin(theta/2)>0#