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Solve the following equation by squaring both sides: $1 + cot(\theta) = csc(\theta)$

$\theta = 0, \pi$

$\theta = 0$

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

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

Explanation

We begin with our original equation:

$1 + cot(\theta) = csc(\theta)$

$(1+cot(\theta))^2 = (csc(\theta))^2$

$1 + 2cot(\theta) + cot^2(\theta) = csc^2(\theta)$

$1 + 2cot(\theta) + cot^2(\theta) = 1 + cot^2(\theta)$ (Pythagorean Identity)

$2cot(\theta) = 0$

$cot(\theta) = 0$

Looking at the unit circle we see that $cot(\theta) = 0$ at $\frac {\pi}{2}$ and $\frac{3\pi}{2}$ . We must plug these back into our original equation to validate them.

Checking $\theta = \frac{\pi}{2}$

$1 + cot(\frac{\pi}{2}) = csc(\frac{\pi}{2})$

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

$1 + cot(\frac{3\pi}{2}) = \frac{3\pi}{2}$

$1 \neq -1$

And so our only solution is $\theta = \frac{\pi}{2}$

If c=70, a=50, and $\angle A=122^\circ$ find $\angle C$ to the nearest degree.

$8.11^\circ$

$49.89^\circ$

$8.11^\circ$ and $171.89^\circ$

$49.89^\circ$ and $130.11^\circ$

no solution

Explanation

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C = 1.18>1$ , which means that there are no solutions to $\angle C$ that satisfy this triangle. If you got answers for this triangle, check that you set up your Law of Sines equation properly at the start of the problem.

While waiting for your sister to finish her bungee jump, you decide to figure out how tall the platform she is jumping off is. You are standing feet from the base of the platform, and the angle of elevation from your position to the top of the platform is degrees. How many feet tall is the platform?

Explanation

You can draw the following right triangle using the information given by the question:

Since you want to find the height of the platform, you will need to use tangent.

$\tan 62=\frac{x}{200}$

Make sure to round to places after the decimal.

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

Solve the following equation by squaring both sides: $1 + cot(\theta) = csc(\theta)$

$\theta = 0, \pi$

$\theta = 0$

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

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

Explanation

We begin with our original equation:

$1 + cot(\theta) = csc(\theta)$

$(1+cot(\theta))^2 = (csc(\theta))^2$

$1 + 2cot(\theta) + cot^2(\theta) = csc^2(\theta)$

$1 + 2cot(\theta) + cot^2(\theta) = 1 + cot^2(\theta)$ (Pythagorean Identity)

$2cot(\theta) = 0$

$cot(\theta) = 0$

Looking at the unit circle we see that $cot(\theta) = 0$ at $\frac {\pi}{2}$ and $\frac{3\pi}{2}$ . We must plug these back into our original equation to validate them.

Checking $\theta = \frac{\pi}{2}$

$1 + cot(\frac{\pi}{2}) = csc(\frac{\pi}{2})$

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

$1 + cot(\frac{3\pi}{2}) = \frac{3\pi}{2}$

$1 \neq -1$

And so our only solution is $\theta = \frac{\pi}{2}$

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.

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

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,

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

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

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

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 $\frac{180^{\circ}}{\pi}$ :

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

What is the domain of f(x) = sin x?

All positive numbers and 0

All negative numbers and 0

All real numbers except 0

All real numbers

Explanation

The domain of a function is the range of all possible inputs, or x-values, that yield a real value for f(x). Trigonometric functions are equal to 0, 1, -1 or undefined when the angle lies on an axis, meaning that the angle is equal to 0, 90, 180 or 270 degrees (0, (pi)/2, pi or 3(pi)/2 in radians.) Trigonometric functions are undefined when they represent fractions with denominators equal to zero. Sine is defined as the ratio between the side length opposite to the angle in question and the hypotenuse (SOH, or sin x = opposite/hypotenuse). In any triangle created by the angle x and the x-axis, the hypotenuse is a nonzero number. As a result, the denominator of the fraction created by the definition sin x = opposite/hypotenuse is not equal to zero for any angle value x. Therefore, the domain of f(x) = sin x is all real numbers.

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