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 $" " "color(red)((1))$

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

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