Pre-Calculus - Find the value of any of the six trigonometric functions

Find the value of $sec(\pi)+cos(-\pi)$ , if possible.

$\frac{\pi^2+1}{\pi}$

$\textup{The answer is not defined.}$

$\frac{1}{2}$

Explanation

In order to solve $sec(\pi)+cos(-\pi)$ , split up the expression into 2 parts.

$sec(\pi)= \frac{1}{cos(\pi)}=\frac{1}{-1}=-1$

$cos(-\pi)= -1$

$sec(\pi)+cos(-\pi)= -1-1=-2$

Simplify the following expression:

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

$\frac{-\sqrt2}{2}$

Explanation

Simplify the following expression:

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

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 $270^{\circ}$

So, we know that

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

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

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

So therefore, our answer must be 1

Simplify the following expression:

$\sec(-2\pi)$

Explanation

Simplify the following expression:

$sec(-2\pi)$

I would begin here by recalling that secant is the reciprocal of cosine. Therefore, we can take the cosine of the given angle and then find its reciprocal.

So,

$cos(-2\pi)=1$

(Because cosine refers to x-values and $-2\pi$ lies on the x-axis)

Therefore,

$sec(-2\pi)=1$

Because $\frac{1}1{}=1$ .

Determine $csc(\frac{3\pi}{4})$

$\sqrt 2$

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

$\frac{1}{\sqrt 2}$

Explanation

Remember that:

$csc(\frac{3\pi}{4})=\frac{1}{sin(\frac{3\pi}{4})}=\frac{1}{\frac{1}{\sqrt 2}}=\sqrt 2$

Q2 new

Find the value of .

Explanation

Using trigonometric relationships, one can set up the equation

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

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

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

Solving for ,

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

Thus, the answer is found to be 106.

Find all of the angles that satistfy the following equation:

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

$\theta= \frac{\pi}{3}+2\pi k$ OR $\theta= \frac{2\pi}{3}+2\pi k$

$\theta= \frac{\pi k}{3}$

$\theta= \frac{\pi}{3},\frac{2\pi}{3}$

$\theta= \frac{2\pi}{3}+2\pi k$

$\theta= \frac{\pi}{3}+\pi k$

Explanation

The values of $\theta$ that fit this equation would be:

$\frac{\pi}{3}$ and $\frac{2\pi}{3}$

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

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

This is why the answer

$\theta= \frac{\pi k}{3}$

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

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

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

$\theta= \frac{\pi}{3}+2\pi k$ OR $\theta= \frac{2\pi}{3}+2\pi k$

What is the value of $tan(\pi)+sin(\pi)$ ?

$\frac{\sqrt2}{2}$

$\infty$

Explanation

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

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

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