Rewrite \(\displaystyle cot\left(\frac{\pi}{2}\right)+tan(\pi)-sec\left(\frac{\pi}{3}\right)\) in terms of sine and cosine functions.

\(\displaystyle \frac{cos(\frac{\pi}{2})}{sin(\frac{\pi}{2})}+\frac{sin(\pi)}{cos(\pi)}-\frac{1}{cos(\frac{\pi}{3})}\)

Since these angles are special angles from the unit circle, the values of each term can be determined from the x and y coordinate points at the specified angle.  

Solve each term and simplify the expression.

\(\displaystyle \frac{0}{1}+\frac{0}{-1}-\frac{1}{\frac{1}{2}}= 0+0-2=-2\)

Using trigonometric relationships, one can set up the equation

\(\displaystyle cos(\theta)=\frac{adjacent}{hypotenuse}\)

\(\displaystyle cos(30^\circ)=\frac{25}{x}\).

Solving for \(\displaystyle x\),

\(\displaystyle \\x\cdot cos(30^\circ)=25\\ \\x=\frac{25}{cos(30^\circ)}\\ \\x=28.8675... \\ \\x=29\)

Thus, the answer is found to be 29.

Using trigonometric relationships, one can set up the equation

\(\displaystyle sin(\theta)=\frac{opposite}{hypotenuse}\).

Plugging in the values given in the picture we get the equation,

\(\displaystyle sin(25) = \frac{45}{x}\).

Solving for \(\displaystyle x\),

\(\displaystyle \\x\cdot sin(25)=45\\ \\x=\frac{45}{sin(25)}\\ \\x=106.479...\\ \\x=106\).

Thus, the answer is found to be 106.

The values of \(\displaystyle \theta\) that fit this equation would be:

\(\displaystyle \frac{\pi}{3}\) and \(\displaystyle \frac{2\pi}{3}\) 

because these angles are in QI and QII where sin is positive and where

\(\displaystyle sin(\theta)=\frac{\sqrt{3}}{2}\).

This is why the answer 

\(\displaystyle \theta= \frac{\pi k}{3}\)

is incorrect, because it includes inputs that provide negative values such as:

\(\displaystyle sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}\)

Thus the answer would be each \(\displaystyle 2\pi\) multiple of \(\displaystyle \frac{\pi}{3}\) and \(\displaystyle \frac{2\pi}{3}\) , which would provide the following equations:

\(\displaystyle \theta= \frac{\pi}{3}+2\pi k\)  OR    \(\displaystyle \theta= \frac{2\pi}{3}+2\pi k\)

To evaluate \(\displaystyle sin(\frac{\pi}{2})-sin(\frac{\pi}{3})+sin(\frac{\pi}{4})\), break up each term into 3 parts and evaluate each term individually.

\(\displaystyle sin(\frac{\pi}{2})=1\)

\(\displaystyle sin(\frac{\pi}{3})=\frac{\sqrt3}{2}\)

\(\displaystyle sin(\frac{\pi}{4})=\frac{\sqrt2}{2}\)

Simplify by combining the three terms.

\(\displaystyle 1-\frac{\sqrt3}{2}+\frac{\sqrt2}{2}= \frac{2-\sqrt3+\sqrt2}{2}\) 

Convert \(\displaystyle tan(\pi)+sin(\pi)\) in terms of sine and cosine.

\(\displaystyle tan(\pi)+sin(\pi)= \frac{sin(\pi)}{cos(\pi)}+sin(\pi)\)

Since theta is \(\displaystyle \pi\) radians, the value of \(\displaystyle sin(\pi)\) is the y-value of the point on the unit circle at \(\displaystyle \pi\) radians, and the value of \(\displaystyle cos(\pi)\) corresponds to the x-value at that angle.

The point on the unit circle at \(\displaystyle \pi\) radians is \(\displaystyle (-1,0)\).  

Therefore, \(\displaystyle sin(\pi)=0\) and \(\displaystyle cos(\pi)=-1\).  Substitute these values and solve.

\(\displaystyle \frac{sin(\pi)}{cos(\pi)}+sin(\pi)=\frac{0}{-1}+0=0\)

First, solve the value of \(\displaystyle sin\left(\frac{\pi}{4}\right)\).  

On the unit circle, the coordinate at \(\displaystyle \frac{\pi}{4}\) radians is \(\displaystyle \left(\frac{\sqrt2}{2}, \frac{\sqrt2}{2}\right)\).  The sine value is the y-value, which is \(\displaystyle \frac{\sqrt2}{2}\).  Substitute this value back into the original problem.

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

Rationalize the denominator.

\(\displaystyle \frac{2}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}= \frac{2\sqrt2}{2}= \sqrt2\)

To evaluate \(\displaystyle cos(60)-sin(30)+cos(30)\), solve each term individually.

\(\displaystyle cos(60)\) refers to the x-value of the coordinate at 60 degrees from the origin.  The x-value of this special angle is \(\displaystyle \frac{1}{2}\).

\(\displaystyle sin(30)\) refers to the y-value of the coordinate at 30 degrees.  The y-value of this special angle is \(\displaystyle \frac{1}{2}\).

\(\displaystyle cos(30)\) refers to the x-value of the coordinate at 30 degrees.  The x-value is \(\displaystyle \frac{\sqrt3}{2}\).

Combine the terms to solve \(\displaystyle cos(60)-sin(30)+cos(30)\).

\(\displaystyle \frac{1}{2}-\frac{1}{2}+\frac{\sqrt3}{2}=\frac{\sqrt3}{2}\)

The value of \(\displaystyle sin(300)\) refers to the y-value of the coordinate that is located in the fourth quadrant.

This angle \(\displaystyle 300\) is also \(\displaystyle -60\) from the origin.  

Therefore, we are evaluating \(\displaystyle sin(-60)\).

\(\displaystyle sin(-60)=-sin(60)=-\frac{\sqrt3}{2}\)

Simplify the following expression:

\(\displaystyle sin(-270^{\circ})\)

Begin by locating the angle on the unit circle. -270 should lie on the same location as 90. We get there by starting at 0 and rotating clockwise \(\displaystyle 270^{\circ}\)

So, we know that 

\(\displaystyle sin(-270^{\circ})=sin(90^{\circ})\)

And since we know that sin refers to y-values, we know that 

\(\displaystyle sin(90^{\circ})=1\)

So therefore, our answer must be 1