Questions: Use a half-angle identity cos(x/2)=±sqrt((1+cos x)/2) to find the exact value of cos(17π/12) a) (sqrt(2-sqrt(3)))/2 b) (sqrt(2+sqrt(3)))/2 c) (sqrt(2+sqrt(3)))/2 d) (sqrt(2+sqrt(3)))/2

Solution Steps

We need to find the exact value of \(\cos\left(\frac{17\pi}{12}\right)\) using the half-angle identity. First, identify the angle \(x\) such that \(\frac{x}{2} = \frac{17\pi}{12}\). Solving for \(x\), we have:

\[ x = 2 \times \frac{17\pi}{12} = \frac{34\pi}{12} = \frac{17\pi}{6} \]

The half-angle identity for cosine is:

\[ \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1+\cos x}{2}} \]

We need to find \(\cos\left(\frac{17\pi}{6}\right)\).

The angle \(\frac{17\pi}{6}\) is equivalent to \(\frac{17\pi}{6} - 2\pi = \frac{17\pi}{6} - \frac{12\pi}{6} = \frac{5\pi}{6}\).

The cosine of \(\frac{5\pi}{6}\) is:

\[ \cos\left(\frac{5\pi}{6}\right) = -\cos\left(\pi - \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \]

Substitute \(\cos\left(\frac{17\pi}{6}\right) = -\frac{\sqrt{3}}{2}\) into the half-angle identity:

\[ \cos\left(\frac{17\pi}{12}\right) = \pm\sqrt{\frac{1 + \left(-\frac{\sqrt{3}}{2}\right)}{2}} = \pm\sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} \]

Simplify the expression:

\[ = \pm\sqrt{\frac{2 - \sqrt{3}}{4}} = \pm\frac{\sqrt{2 - \sqrt{3}}}{2} \]

Since \(\frac{17\pi}{12}\) is in the second quadrant, where cosine is negative, we choose the negative sign:

\[ \cos\left(\frac{17\pi}{12}\right) = -\frac{\sqrt{2 - \sqrt{3}}}{2} \]

Final Answer