suppose that $ \sec \alpha = \frac { 17 } { 15 } $ and $ 0 ^ { \circ }

Question

suppose that $ \sec \alpha = \frac { 17 } { 15 } $ and $ 0 ^ { \circ } < \alpha < 90 ^ { \circ } $.
find the exact values of $ \cos \frac { \alpha } { 2 } $ and $ \tan \frac { \alpha } { 2 } $.
$ \cos \frac { \alpha } { 2 } = $
$ \tan \frac { \alpha } { 2 } = $

Accepted Answer

# Explanation:
## Step1: Find $\cos\alpha$
Since $\sec\alpha=\frac{17}{15}$, and $\sec\alpha=\frac{1}{\cos\alpha}$, then $\cos\alpha=\frac{15}{17}$.
## Step2: Use the half - angle formula for $\cos\frac{\alpha}{2}$
The half - angle formula is $\cos\frac{\alpha}{2}=\sqrt{\frac{1 + \cos\alpha}{2}}$.
Substitute $\cos\alpha=\frac{15}{17}$ into the formula:
$$
\begin{align*}
\cos\frac{\alpha}{2}&=\sqrt{\frac{1+\frac{15}{17}}{2}}\
&=\sqrt{\frac{\frac{17 + 15}{17}}{2}}\
&=\sqrt{\frac{32}{17\times2}}\
&=\sqrt{\frac{16}{17}}\
&=\frac{4}{\sqrt{17}}\
&=\frac{4\sqrt{17}}{17}
\end{align*}
$$
## Step3: Use the half - angle formula for $\tan\frac{\alpha}{2}$
The half - angle formula is $\tan\frac{\alpha}{2}=\sqrt{\frac{1-\cos\alpha}{1 + \cos\alpha}}$.
Substitute $\cos\alpha=\frac{15}{17}$ into the formula:
$$
\begin{align*}
\tan\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{15}{17}}{1+\frac{15}{17}}}\
&=\sqrt{\frac{\frac{17-15}{17}}{\frac{17 + 15}{17}}}\
&=\sqrt{\frac{2}{32}}\
&=\sqrt{\frac{1}{16}}\
&=\frac{1}{4}
\end{align*}
$$

# Answer:
$\cos\frac{\alpha}{2}=\frac{4\sqrt{17}}{17}$, $\tan\frac{\alpha}{2}=\frac{1}{4}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Find \(\cos\alpha\)

Step2: Use the half - angle formula for \(\cos\frac{\alpha}{2}\)

Step3: Use the half - angle formula for \(\tan\frac{\alpha}{2}\)

Answer: