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

1 Answer
Jun 23, 2015

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

Explanation:

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#