How do you find sin (x/2), given $cos x = -5/8$ , with $pi/2 < x <pi$ ?

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Answer 1 · 2015-05-18T14:34:17

Recall the formula for a cosine of a sum of two angles:
$cos(alpha+beta)=cos(alpha)*cos(beta)-sin(alpha)*sin(beta)$

Use it for $alpha=beta=x/2$ :
$cos(x/2+x/2)=cos^2(x/2)-sin^2(x/2)$

Substitute $x/2+x/2=x$
Use the identity $sin^2(gamma)+cos^2(gamma)=1$ for $gamma=x/2$
The result is:
$cos(x)=1-2sin^2(x/2)$

Now we can use the value of $cos(x)=-5/8$ .
$1-2sin^2(x/2)=-5/8$
$sin^2(x/2)=(1+5/8)/2=13/16$
$sin(x/2)=+-sqrt(13)/4$

If $pi/2 < x < pi$ , $pi/4 < x/2 < pi/2$ . For these angles function sine is positive.
Therefore, $sin(x/2)=sqrt(13)/4$ .

How do you find sin (x/2), given cos x = -5/8cos x = -5/8, with pi/2 < x <pipi/2 < x <pi?