How do you evaluate #sin(cos^-1(1/2))# without a calculator?

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Answer 1 · 2018-03-05T14:59:40

#sin(cos^(-1)(1/2))=sqrt(3)/2#

Let #cos^(-1)(1/2)=x# then #cosx=1/2#

#rarrsinx=sqrt(1-cos^2x)=sqrt(1-(1/2)^2)=sqrt(3)/2#

#rarrx=sin^(-1)(sqrt(3)/2)=cos^(-1)(1/2)#

Now, #sin(cos^(-1)(1/2))=sin(sin^(-1)(sqrt(3)/2))=sqrt(3)/2#

Answer 2 · 2018-03-05T15:13:41

#sin cos ^-1 (1/2)) = sqrt 3 / 2#

To find value of #sin (cos ^-1 (1/2))#

Let theta = cos^-1 (1/2)#

#cos theta = (1/2)#

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We know, from the above table, #cos 60 = 1/2#

Hence theta = 60^@#

Replacing #cos^-1 (1/2)# with #theta = 60^@#,

The sum becomes, #=> sin theta = sin 60 = sqrt3 / 2# (As per table above)