Trigonometry

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Which of the following is the correct definition of a phase shift?

A measure of the length of a function between vertical asymptotes

The distance a function is shifted diagonally from the general position

The distance a function is shifted horizontally from the general position

The distance a function is shifted vertically from the general position

Explanation

Take the function for example. The graph for is

If we were to change the function to $cos(x + \pi)$ , our phase shift is $-\pi$ . This means we need to shift our entire graph $\pi$ units to the left.

Our new graph $cos(x + \pi)$ is the following

Change the following expression to degrees:

$\frac{\frac{\pi}{3}-\frac{\pi}{6}}{\frac{1}{3}+\frac{1}{6}}$

$60^{\circ}$

$90^{\circ}$

$75^{\circ}$

$160^{\circ}$

$120^{\circ}$

Explanation

First we need to simplify the expression:

$\frac{\frac{\pi}{3}-\frac{\pi}{6}}{\frac{1}{3}+\frac{1}{6}}=\frac{\frac{2\pi-\pi}{6}}{\frac{2+1}{6}}=\frac{\pi}{3}$

Now multiply by $\frac{180^{\circ}}{\pi}$ :

$\frac{\pi}{3}\times \frac{180^{\circ}}{\pi}=\frac{180^{\circ}}{3}=60^{\circ}$

Convert $\frac{2\pi}{5}$ to degrees:

$72^{\circ}$

$114^{\circ}$

$36^{\circ}$

$\frac{2}{5}^{\circ}$

$72\pi$

Explanation

To convert radians to degrees, we need to multiply the given radians by $\frac{180^{\circ}}{\pi}$ .

$\frac{2\pi}{5}*\frac{180^{\circ}}{\pi}=\frac{360^{\circ}}{5}=72^{\circ}$

Convert $\pi$ into degrees.

$180^{\circ}$

$90^{\circ}$

$30^{\circ}$

$360^{\circ}$

$3.14^{\circ}$

Explanation

Recall that there are 360 degrees in a circle which is equivalent to $2\pi$ radians. In order to convert between radians and degrees use the relationship that,

$360^\circ=2\pi \Rightarrow 180^\circ=\pi$ .

Therefore, in order to convert from radians to degrees you need to multiply by $\frac{180}{\pi}$ . So in this particular case,

$\pi*\frac{180}{\pi}=180^{\circ}$ .

What angle is complementary to $34^\circ$ ?

$56^\circ$

$66^\circ$

$34^\circ$

$146^\circ$

$156^\circ$

Explanation

Two complementary angles add up to $90^\circ$ .

Therefore, $34^\circ+x^\circ=90^\circ$ .

$x^\circ=90^\circ-34^\circ$

$x^\circ=56^\circ$

Trig_id

What is $\theta$ if and ?

$51.34^{\circ}$

$53.14^{\circ}$

$55.14^{\circ}$

$0.90^{\circ}$

$89.60^{\circ}$

Explanation

In order to find $\theta$ we need to utilize the given information in the problem. We are given the opposite and adjacent sides. We can then, by definition, find the $\tan$ of $\theta$ and its measure in degrees by utilizing the $\arctan$ function.

$\tan=\frac{opposite}{adjacent}$

$\tan=\frac{10}{8}$

$\tan=1.25$

Now to find the measure of the angle using the $\arctan$ function.

$\theta=\arctan1.25$

$\rightarrow 51.34^{\circ}$

If you calculated the angle's measure to be $0.90^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

Trig_id

What is $\theta$ if and ?

$51.34^{\circ}$

$53.14^{\circ}$

$55.14^{\circ}$

$0.90^{\circ}$

$89.60^{\circ}$

Explanation

$\tan=\frac{opposite}{adjacent}$

$\tan=\frac{10}{8}$

$\tan=1.25$

Now to find the measure of the angle using the $\arctan$ function.

$\theta=\arctan1.25$

$\rightarrow 51.34^{\circ}$

If you calculated the angle's measure to be $0.90^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

Which of these is equal to $\frac{\text{hypotenuse}}{\text{adjacent}}$ for angle ?

$\sec{x}$

$\csc{x}$

$\sin x$

$\cot{x}$

$\cos{x}$

Explanation

$\sec{x}=\frac{\text{hypotenuse}}{\text{adjacent}}$ , as it is the inverse of the $\cos{x}$ function. This is therefore the answer.

Using trigonometric identities prove whether the following is valid:

$\sin^{2}A + \cot^{2}A = \frac{2\sin^{4}A +2\cos^{2}A}{1 - \cos 2A}$

True

False

Uncertain

Only in the range of: $0 < A < \frac{\pi}{2}$

Only in the range of: $0 < A < 2\pi$

Explanation

We can work with either side of the equation as we choose. We work with the right hand side of the equation since there is an obvious double angle here. We can factor the numerator to receive the following:

$\sin^{2}A + \cot^{2}A = \frac{2\left(\sin^{4}A + \cos^{2}A\right)}{1-\cos 2A}$

Next we note the power reducing formula for sine so we can extract the necessary components as follows:

$\sin^{2}A + \cot^{2}A = \frac{2}{1-\cos 2A}\left(\sin^{4}A + \cos^{2}A\right)$

The power reducing formula must be inverted giving:

$\sin^{2}A + \cot^{2}A = \frac{1}{\sin^{2}A}\left(\sin^{4}A + \cos^{2}A\right)$

Now we can distribute and reduce:

$\sin^{2}A + \cot^{2}A = \sin^{2}A + \frac{\cos^{2}A}{\sin^{2}A}$

Finally recalling the basic identity for the cotangent:

$\sin^{2}A + \cot^{2}A = \sin^{2}A + \cot^{2}A$

This proves the equivalence.

Simplify

$\frac{\sin^{2}{\left (x \right )} + \cos^{3}{\left (x \right )}}{2 \cos^{2}{\left (x \right )} - 1}$

$\frac{1}{2} \left(\cos^{3}{\left (x \right )} \csc^{2}{\left (x \right )} + 1\right)$

$\frac{1}{2} \cos^{3}{\left (x \right )} \csc^{2}{\left (x \right )}$

$\cos^{2}{\left (x \right )} + \csc^{3}{\left (x \right )}$

$\sin^{2}{\left (x \right )} + \sec^{2}{\left (x \right )}$

$\frac{1}{2}$

Explanation

The first step to simplifying is to remember an important trig identity.

$\sin^{2}{\left (x \right )} + \cos^{2}{\left (x \right )} = 1$

If we rewrite it to look like the denominator, it is.

$\sin^{2}{\left (x \right )} = \cos^{2}{\left (x \right )} - 1$

Now we can substitute this in the denominator.

$\frac{\sin^{2}{\left (x \right )} + \cos^{3}{\left (x \right )}}{2 \sin^{2}{\left (x \right )}}$

Now write each term separately.

$\frac{1}{2} + \frac{\cos^{3}{\left (x \right )}}{2 \sin^{2}{\left (x \right )}}$

Remember the following identities.

$\\csc{\left (x \right )} = \frac{1}{\sin{\left (x \right )}} \\\sec{\left (x \right )} = \frac{1}{\cos{\left (x \right )}} \\\tan{\left (x \right )} = \frac{\sin{\left (x \right )}}{\cos{\left (x \right )}} \\\cot{\left (x \right )} = \frac{\cos{\left (x \right )}}{\sin{\left (x \right )}}$

Now simplify, and combine each term.

$\\frac{1}{2} \cos^{3}{\left (x \right )} \csc^{2}{\left (x \right )} + \frac{1}{2} \\\frac{1}{2} \left(\cos^{3}{\left (x \right )} \csc^{2}{\left (x \right )} + 1\right)$

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