To find \(\cos \left( \frac{t}{2} \right)\) given that \(\frac{3\pi}{2} < t < 2\pi\) and \(\cos(t) = \frac{3}{4}\), we can use the half-angle formula for cosine.
The half-angle formula for cosine states:
[tex]\[
\cos \left( \frac{t}{2} \right) = \pm \sqrt{\frac{1 + \cos(t)}{2}}
\][/tex]
We need to determine the correct sign for \(\cos \left( \frac{t}{2} \right)\). Since \( \frac{3\pi}{2} < t < 2\pi \), dividing these bounds by 2 gives \( \frac{3\pi}{4} < \frac{t}{2} < \pi \).
The cosine function is negative in the interval \( \left( \frac{3\pi}{4}, \pi \right) \). Therefore, \(\cos \left( \frac{t}{2} \right) \) will be negative in this range.
Now, substitute \(\cos(t) = \frac{3}{4}\) into the half-angle formula:
[tex]\[
\cos \left( \frac{t}{2} \right) = - \sqrt{\frac{1 + \frac{3}{4}}{2}}
\][/tex]
Simplify the expression inside the square root:
[tex]\[
\cos \left( \frac{t}{2} \right) = - \sqrt{\frac{1 + 3/4}{2}} = - \sqrt{\frac{4/4 + 3/4}{2}} = - \sqrt{\frac{7/4}{2}} = - \sqrt{\frac{7/4}{2/1}} = - \sqrt{\frac{7/4 \cdot 1/2}} = - \sqrt{\frac{7}{8}}
\][/tex]
Simplify the square root:
[tex]\[
\cos \left( \frac{t}{2} \right) = - \sqrt{\frac{7}{8}} = - \frac{\sqrt{7}}{\sqrt{8}} = - \frac{\sqrt{7}}{2\sqrt{2}} = - \frac{\sqrt{7}}{2} \cdot \frac{1}{\sqrt{2}} = - \frac{\sqrt{7}}{2} \cdot \frac{\sqrt{2}}{2} = - \frac{\sqrt{7 \cdot 2}}{4} = - \frac{\sqrt{14}}{4}
\][/tex]
Thus, the exact value of \(\cos \left( \frac{t}{2} \right)\) is:
[tex]\[
\cos \left( \frac{t}{2} \right) = - \frac{\sqrt{14}}{4}
\][/tex]
