To derive the graph of \(\displaystyle cot(x)\) recall that \(\displaystyle cot(x) = \frac{1}{tan(x)}\).  So the tangent and cotangent graphs are reciprocals of one another.  We will consider the tangent graph since it is one we are more familiar with:

Now we will simply invert the tangent graph to get the cotangent graph

And we are left with our cotangent graph

First, we will consider the graph of \(\displaystyle cot(x)\) and apply transformations step-by-step.  The graph of \(\displaystyle cot(x)\) is

The general form of a cotangent transformation function is \(\displaystyle D + Acot(B(x - C))\).  For our function \(\displaystyle 4cot(x + \frac{2\pi}{5})\), \(\displaystyle A = 4\) and so we need to increase the amplitude 4 units.

\(\displaystyle B = 1\) here so we do not need to make any changes to the period of this graph.  \(\displaystyle C = -\frac{2\pi}{5}\), giving us a negative phase shift of \(\displaystyle \frac{2\pi}{5}\) units.

This leaves us with our graph of \(\displaystyle 4cot(x + \frac{2\pi}{5})\).

This is because \(\displaystyle tan(x) = \frac{sin(x)}{cos(x)}\) and \(\displaystyle cos(n\pi) = 0\)  causing the tangent function to be undefined at these points and forming a vertical asymptote.  This is also true for the cotangent function because \(\displaystyle cot(x) = \frac{1}{tan(x)}\) so wherever \(\displaystyle tan(x)\) is zero or undefined, cotangent will be as well.

This is the figure being described in the problem, and as sine is opposite over hypotenuse and cosine adjacent over hypotenuse, the sine of \(\displaystyle a\) and the cosine of \(\displaystyle b\) will yield the correct answer.

\(\displaystyle \sin(a)=\frac{opp}{hyp}=\frac{8}{17}\)

\(\displaystyle \cos(b)=\frac{adj}{hyp}=\frac{8}{17}\)

Rearrange algebraically so that,

\(\displaystyle 3sin(x)=3\)

 \(\displaystyle sinx = 1\).

Over the interval 0 to 360 degrees, sinx = 1 when x equals 90 degrees.

First, think of the angle values for which \(\displaystyle tanx = 1\). (This is the equivalent of taking the arctan.)

\(\displaystyle x-15=tan^{-1}(1)\)

\(\displaystyle x=tan^{-1}+15\).

The angles for which this is true are 45 degrees and 225 degrees.

We set x-15 equal to those two angles and solve for x, giving us 60 and 240. 

Rearrange the equation so that,

 \(\displaystyle sec(x+7) = 2\).

Recall the angles over the interval 0 to 360 degrees for which sec is equal to 2.

These are 60 and 300 degrees.

Set x+7 equal to these angle measures and then find that x equals 53 and 293. 

Rearrange the equation so that,

 \(\displaystyle cos^2x = \frac{1}{2}\).

Take the square of both sides and then recall the angle measures for which,

\(\displaystyle cos(x)={\frac{1}{\sqrt2}}\).

These measures over the interval \(\displaystyle 0< x< 2 \pi\)

 \(\displaystyle x=\frac{\pi}{4}; \frac{7\pi}{4}\).

Rearrange the equation so that,

\(\displaystyle sin^2x =1\).

Take the square of both sides and find the angles for which

\(\displaystyle sin(x)=\pm1\).

These two angles are \(\displaystyle \frac{\pi}{2}\) and \(\displaystyle \frac{3\pi}{2}\).

To solve for x, first set up a trigonometric equation using the information provided in the diagram. The two side lengths given are the hypotenuse, x, and the side opposite the given angle, 6. We can set up our equation like this:

\(\displaystyle sin\left(\frac{\pi}{4}\right)= \frac{6}{x}\)

The sine of \(\displaystyle \frac{\pi}{4}\) is \(\displaystyle \frac{1}{\sqrt2}\), so we can substitue that in:

\(\displaystyle \frac{1}{\sqrt2}=\frac{6}{x}\)

cross multiplying gives us \(\displaystyle x=6\sqrt2\).