Question

find the ( cos \frac{5pi}{8} ) using the half-angle formula. rationalize the denominator if necessary.
half-angle formulas
( sinleft(\frac{\theta}{2}
ight) = pm sqrt{\frac{1 - cos(\theta)}{2}} )
( cosleft(\frac{\theta}{2}
ight) = pm sqrt{\frac{1 + cos(\theta)}{2}} )
( cos \frac{5pi}{8} = -\frac{sqrt{?} - sqrt{square}}{} )

Explanation:

Step1: Identify θ for half - angle

We know that \(\cos\frac{5\pi}{8}=\cos\frac{\frac{5\pi}{4}}{2}\). So, let \(\theta = \frac{5\pi}{4}\) and we are using the half - angle formula \(\cos\frac{\theta}{2}=\pm\sqrt{\frac{1 + \cos\theta}{2}}\). Now, we need to determine the sign of \(\cos\frac{5\pi}{8}\). The angle \(\frac{5\pi}{8}\) is in the second quadrant (\(\frac{\pi}{2}<\frac{5\pi}{8}<\pi\)), and in the second quadrant, cosine is negative. So, \(\cos\frac{5\pi}{8}=-\sqrt{\frac{1+\cos\frac{5\pi}{4}}{2}}\).

Step2: Calculate \(\cos\frac{5\pi}{4}\)

We know that \(\cos\frac{5\pi}{4}=\cos(\pi+\frac{\pi}{4})=-\cos\frac{\pi}{4}=-\frac{\sqrt{2}}{2}\) (using the identity \(\cos(A + B)=\cos A\cos B-\sin A\sin B\), when \(A=\pi\) and \(B = \frac{\pi}{4}\), \(\cos(\pi+\frac{\pi}{4})=\cos\pi\cos\frac{\pi}{4}-\sin\pi\sin\frac{\pi}{4}=(- 1)\times\frac{\sqrt{2}}{2}-0\times\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{2}\)).

Step3: Substitute \(\cos\frac{5\pi}{4}\) into the formula

Substitute \(\cos\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}\) into \(\cos\frac{5\pi}{8}=-\sqrt{\frac{1+\cos\frac{5\pi}{4}}{2}}\):
\[

$$\begin{align*} \cos\frac{5\pi}{8}&=-\sqrt{\frac{1+(-\frac{\sqrt{2}}{2})}{2}}\\ &=-\sqrt{\frac{\frac{2-\sqrt{2}}{2}}{2}}\\ &=-\sqrt{\frac{2 - \sqrt{2}}{4}}\\ &=-\frac{\sqrt{2-\sqrt{2}}}{2} \end{align*}$$

\]
To match the given form \(-\frac{\sqrt{[?]-\sqrt{\square}}}{2}\), we can see that the first square (the one inside the outer square root) is \(2\) and the second square is \(2\) (since \(\sqrt{2-\sqrt{2}}\) can be written as \(\sqrt{2}-\sqrt{2}\) in the numerator of the fraction under the square root when we rationalize or analyze the form). Wait, actually, let's re - express the steps to fit the form.

We have \(\cos\frac{5\pi}{8}=-\sqrt{\frac{1+\cos\frac{5\pi}{4}}{2}}\), \(\cos\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}\), so:

\[

$$\begin{align*} \cos\frac{5\pi}{8}&=-\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\\ &=-\sqrt{\frac{2 - \sqrt{2}}{4}}\\ &=-\frac{\sqrt{2-\sqrt{2}}}{2} \end{align*}$$

\]

Comparing with the form \(-\frac{\sqrt{[?]-\sqrt{\square}}}{2}\), we can see that \([?] = 2\) and \(\square=2\).

Answer:

The value of \(\cos\frac{5\pi}{8}\) is \(-\frac{\sqrt{2-\sqrt{2}}}{2}\), so in the given form, the first blank (inside the square root) is \(2\) and the second blank is \(2\). So the answer is \(-\frac{\sqrt{2-\sqrt{2}}}{2}\) (or in the boxed form as per the problem's requirement for the blanks, the first blank is \(2\) and the second blank is \(2\)).