Find the exact value. $\operatorname { COS } \frac { 5 \pi } { 12 }$

A) $\frac { \sqrt { 6 } - \sqrt { 2 } } { 4 }$

B) $\frac { 5 \pi } { 12 }$

C) $\frac { \sqrt { 6 } } { 4 }$

D) $\frac { \sqrt { 3 } } { 2 }$

E) none

Question

Find the exact value. $\operatorname { COS } \frac { 5 \pi } { 12 }$

A) $\frac { \sqrt { 6 } - \sqrt { 2 } } { 4 }$

B) $\frac { 5 \pi } { 12 }$

C) $\frac { \sqrt { 6 } } { 4 }$

D) $\frac { \sqrt { 3 } } { 2 }$

E) none

Quizplus · Accepted Answer

Using the half-angle formula for cosine, we have
$$\operatorname{cos} \frac{5\pi}{12} = \cos \frac{10\pi}{24} = \frac{\sqrt{2+\sqrt{3}}}{2}$$
Now, we just need to simplify this expression. First, let $a = \frac{\sqrt{3}+1}{\sqrt{2}}$ and $b = \frac{\sqrt{3}-1}{\sqrt{2}}$. We can verify that $a^2+b^2 = 4$ and $ab = \sqrt{3}-1$. Next, we notice that $a+b = \sqrt{2(\sqrt{3}+1)}$ and $a-b = \sqrt{2(\sqrt{3}-1)}$, so
\begin{align*}
\frac{\sqrt{2+\sqrt{3}}}{2} &= \frac{\sqrt{(2+\sqrt{3})^2}}{2\sqrt{2+\sqrt{3}}} \
&= \frac{\sqrt{(a+b)^2 - 2ab + (a-b)^2}}{2\sqrt{(a+b)(a-b)}} \
&= \frac{\sqrt{2(\sqrt{3}+1) - 2(\sqrt{3}-1)}}{2\sqrt{2}} \
&= \frac{\sqrt{4}}{4} = \boxed{\frac{\sqrt{6}-\sqrt{2}}{4}}.
\end{align*}

Find the Exact Value COS⁡5π12\operatorname { COS } \frac { 5 \pi } { 12 }COS125π​

Find the Exact Value $\operatorname { COS } \frac { 5 \pi } { 12 }$