The exact value of \(\displaystyle cos(45)\) is the x-value when the angle is 45 degrees on the unit circle.  

The x-value of this angle is \(\displaystyle \frac{\sqrt2}{2}\).

\(\displaystyle 8cos(45)= 8\left(\frac{\sqrt{2}}{2}\right)=4\sqrt2\)

Remember SOH-CAH-TOA! That means:

\(\displaystyle \small cos(x)=\frac{adjacent}{hypotenuse}=\frac{b}{c}\)

    \(\displaystyle \small \small \small sin(x)=\frac{a}{c}\)                \(\displaystyle \small \small \small \small \small sin(y)=\frac{b}{c}\)  

                                     \(\displaystyle \small \small \small \small cos(y)=\frac{a}{c}\)  

    \(\displaystyle \small \small \small \small \small tan(x)=\frac{a}{b}\)                \(\displaystyle \small \small \small \small tan(y)=\frac{b}{a}\)  

sin(y) is equal to cos(x)

To find the value of \(\displaystyle \frac{1}{2}sin(45)+ tan(60)\), solve each term separately.

\(\displaystyle \frac{1}{2}sin(45)=\frac{1}{2} \cdot \frac{\sqrt2}{2} = \frac{\sqrt2}{4}\)

\(\displaystyle tan(60) = \sqrt3\)

Sum the two terms.

\(\displaystyle \frac{\sqrt2}{4}+\sqrt3 = \frac{\sqrt2}{4}+\frac{4\sqrt3}{4} = \frac{\sqrt2+4\sqrt3}{4}\)

The tangent function has a period of \(\displaystyle \pi\) units. That is,

\(\displaystyle tan(x+n\pi)=tanx\)

for all \(\displaystyle n\in\mathbb{Z}\).

Since \(\displaystyle \frac{4\pi}{3}=\frac{\pi}{3}+\pi\), we can rewrite the original expression \(\displaystyle tan(\frac{4\pi}{3})\) as follows:

\(\displaystyle tan(\frac{4\pi}{3})=tan(\frac{\pi}{3}+\pi)\)

                 \(\displaystyle =tan(\frac{\pi}{3})\)

                 \(\displaystyle =\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})}\)

                 \(\displaystyle =\frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})}\)

                 \(\displaystyle =\sqrt{3}\)

Hence, 

\(\displaystyle tan(\frac{4\pi}{3})=\sqrt{3}\)

First, convert the given angle measure from radians to degrees:

\(\displaystyle cos(\frac{5\pi}{3})=cos(300^{\circ})\)

Next, recall that \(\displaystyle 300^{\circ}\) lies in the fourth quadrant of the unit circle, wherein the cosine is positive. Furthermore, the reference angle of \(\displaystyle 300^{\circ}\) is 

\(\displaystyle 360^{\circ}-300^{\circ}=60^{\circ}\)

Hence, all that is required is to recognize from these observations that 

\(\displaystyle cos(\frac{5\pi}{3})=cos(300^{\circ}) =cos(60^{\circ})\),

which is \(\displaystyle \frac{1}{2}\).

Therefore,

\(\displaystyle cos(\frac{5\pi}{3})=\frac{1}{2}\)

Since cotangent is positive and sine is negative, alpha must be in quadrant III.  \(\displaystyle \cot \alpha = \frac{x}{y}\) then implies that \(\displaystyle (x,y)=(-5,-1)\) is a point on the terminal side of alpha. 

\(\displaystyle r=\sqrt{(x^2+y^2)}=\sqrt{26}\)

\(\displaystyle \sec \alpha = \frac{r}{x} = \frac{\sqrt{26}}{-5}\)

Since cosecant is negative, theta must be in quadrant III or IV. 

Since tangent is positive, it must be in quadrant I or III. 

Therefore, theta must be in quadrant III.

Using a unit circle we can see that quadrant III is when theta is between \(\displaystyle \pi\) and \(\displaystyle \frac{3\pi}{2}\).

Secant is defined to be the ratio of \(\displaystyle r\) to \(\displaystyle x\) where \(\displaystyle r\) is the distance from the origin. 

The Pythagoreanr Triple 5, 12, 13 helps us realize that \(\displaystyle r = 13\). 

Since \(\displaystyle x = 12\), the answer is \(\displaystyle \frac{13}{12}\).

Use the following trigonometric identity to solve this problem.

\(\displaystyle \csc (x+y) = \frac {1}{\sin (x+y)} = \frac {1}{\sin x \cos y + \cos x \sin y}\)

Using the Pythagorean triple 3,4,5, it is easy to find \(\displaystyle \cos x = \frac {4}{5}\).

Using the Pythagorean triple 5,12,13, it is easy to find \(\displaystyle \sin y = \frac {12}{13}\).

So substituting all four values into the top equation, we get

\(\displaystyle \csc (x+y) = \frac {1}{\frac {3}{5} \cdot \frac {5}{13} + \frac {4}{5} \cdot \frac {12}{13}} = \frac {65}{63}\)

The value of the secant of an angle is the value of the hypotenuse over the adjacent.

Therefore:

\(\displaystyle sec \angle A = \frac{hypotenuse}{adjacent} = \frac{25}{24}\)