How do you evaluate #sin (arccos (2/8))#?

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Answer 1 · 2015-06-23T08:39:08

#sin(arccos(2/8))#=#sqrt(15)/4#

Let

#sin(cos^(-1)(2/8))=sin(x)##" " "color(red)((1))#

From #color(red)((1))#;

#x=cos^(-1)(2/8)#

#cosx=2/8#

#sinx=sqrt(1-cos^2(x)#

#sinx=sqrt(1-(2/8)^2#

#sinx=sqrt(1-4/64#

#sinx=sqrt(64-4)/sqrt64#

#sinx=sqrt60/sqrt64#

#sinx=sqrt(15xx4)/8#

#sinx=(2sqrt(15))/8#

#sinx=sqrt(15)/4#

The above sinx value substitute in #" " "color(red)((1))#

#sin(arccos(2/8))#=#sqrt(15)/4#