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In the figure below, $\overline{BD}$ is a diagonal of quadrilateral . $\overline{BD}$ has a length of . $\angle CBD$ is congruent to $\angle BDC$ .

Screen shot 2020 08 27 at 4.39.20 pm

Which of the following is a true statement?

The area of quadrilateral is $\frac{1}{2}$ .

The area of quadrilateral is .

The perimeter of quadrilateral is .

Explanation

Since and are perpendicular, $\angle BCD$ is a right angle. Since no triangle can have more than one right angle, and $\Delta BCD$ is isosceles, $\angle CBD$ must be congruent to $\angle BDC$ . Since angle CBD is congruent to $\angle BDC$ and $\angle BCD$ measures 90 degrees, $\angle CBD$ and $\angle BDC$ can be calculated as follows:

Therefore, and are both equal to 45 degrees. $\Delta BCD$ is a 45-45-90 triangle. Therefore, the ratio between side lengths and hypotenuse is $1:\sqrt{2}$ . Anyone of the four side lengths of quadrilateral must, therefore, be equal to $\frac{BD}{\sqrt{2}}=\frac{1}{\sqrt{2}}$ . To find the area of , multiply two side lengths: $\frac{1}{\sqrt{2}}*\frac{1}{\sqrt{2}}=\frac{1}{2}$ .

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

Which of the following is positive?

$\cos 45$

$\cos 90$

$\cos 135$

$\cos 180$

Explanation

When drawn from the origin, a line 45 degrees above (counterclockwise from) the positive x-axis lies in quadrant I. Cosine is defined as the ratio between the adjacent side of a triangle and the hypotenuse of the triangle. A right triangle can be drawn in quadrant I composed of any point on that line, the origin and a point on the x-axis. The hypotenuse of this triangle is considered a length, and is therefore positive. The adjacent side of this triangle lies along the positive x-axis. Since the adjacent side and hypotenuse are both represented by positive numbers, the fraction A/H is positive. Therefore, cos 45 is positive.

You can derive the formula $1+\tan^2\theta=\sec^2\theta$ by dividing the formula $\sin^2\theta+\cos^2\theta=1$ by which of the following functions?

$\cos\theta$

$\sin\theta$

$\cos^2\theta$

$\sin^2\theta$

$\tan^2\theta$

Explanation

The correct answer is $\cos^2\theta$ . Rather than memorizing all three Pythagorean Relationships, you can memorize only $\sin^2\theta+\cos^2\theta=1$ , then simply divide all terms by $\cos^2\theta$ to get the formula that relates $\tan^2\theta$ and $\sec^2\theta$ . Alternatively, you can divide all terms of $\sin^2\theta+\cos^2\theta=1$ by $\sin^2\theta$ to get the formula that relates $\cot^2\theta$ and $\csc^2\theta$ . The former is demonstrated below.

$\sin^2\theta+\cos^2\theta=1$

$\frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}$

$\tan^2\theta+1=\sec^2\theta$

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

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.

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.

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

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.

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?