Circular Functions

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Pre-Calculus › Circular Functions

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Please choose the best answer from the following choices.

Find the secant value of if it's a point on the terminal side of an angle in standard position.

$\sec \theta=\sqrt{97}/-9$

$\sec \theta=\ -9/\sqrt{97}$

$\sec \theta= 4/\sqrt{97}$

$\sec \theta=\sqrt{97}/4$

Explanation

First, use the Pythagorean Theorem to solve for all the sides of the triangle. You know that the adjacent side is 4 units long, and the opposite side is -9 units long.

Using the Pythagorean Theorem, you should get a hypotenuse of $\sqrt{97}$ .

$\a^2+b^2=c^2 \ \ 4^2+(-9)^2=97 \ \c^2=\sqrt{97}$

Secant is defined as hypotenuse/opposite.

Thus, giving you an answer of $\sqrt{97}/-9$ .

Please choose the best answer from the following choices.

Find the secant value of if it's a point on the terminal side of an angle in standard position.

$\sec \theta=\sqrt{97}/-9$

$\sec \theta=\ -9/\sqrt{97}$

$\sec \theta= 4/\sqrt{97}$

$\sec \theta=\sqrt{97}/4$

Explanation

First, use the Pythagorean Theorem to solve for all the sides of the triangle. You know that the adjacent side is 4 units long, and the opposite side is -9 units long.

Using the Pythagorean Theorem, you should get a hypotenuse of $\sqrt{97}$ .

$\a^2+b^2=c^2 \ \ 4^2+(-9)^2=97 \ \c^2=\sqrt{97}$

Secant is defined as hypotenuse/opposite.

Thus, giving you an answer of $\sqrt{97}/-9$ .

What is the sine of an angle if a point on the terminal side of the angle is ?

$\frac{5\sqrt{26}}{26}$

$\frac{\sqrt{26}}{26}$

$\frac{5}{6}$

$\frac{1}{6}$

Explanation

Given the point on the coordinate plane , the origin to this point can be computed by the Pythagorean Theorem.

$c=\sqrt{(1^2+5^2)}=\sqrt{26}$

The hypotenuse of the right triangle formed by the origin and the point is $\sqrt{26}$ .

The length of the triangle is 1 unit, and the height of the triangle is 5.

Sine of an angle is opposite side divided by the hypotenuse.

$sin(\theta)=\frac{\textup{Opposite side}}{\textup{Hypotenuse}}= \frac{5}{\sqrt{26}}$

Rationalize the denominator.

$\frac{5}{\sqrt{26}}\cdot \frac{\sqrt{26}}{\sqrt{26}}= \frac{5\sqrt{26}}{26}$

What is the sine of an angle if a point on the terminal side of the angle is ?

$\frac{5\sqrt{26}}{26}$

$\frac{\sqrt{26}}{26}$

$\frac{5}{6}$

$\frac{1}{6}$

Explanation

Given the point on the coordinate plane , the origin to this point can be computed by the Pythagorean Theorem.

$c=\sqrt{(1^2+5^2)}=\sqrt{26}$

The hypotenuse of the right triangle formed by the origin and the point is $\sqrt{26}$ .

The length of the triangle is 1 unit, and the height of the triangle is 5.

Sine of an angle is opposite side divided by the hypotenuse.

$sin(\theta)=\frac{\textup{Opposite side}}{\textup{Hypotenuse}}= \frac{5}{\sqrt{26}}$

Rationalize the denominator.

$\frac{5}{\sqrt{26}}\cdot \frac{\sqrt{26}}{\sqrt{26}}= \frac{5\sqrt{26}}{26}$

Find the sine and cosine of the following angle.

Screen shot 2020 08 21 at 4.56.27 pm

$\sin = \frac{6\sqrt{59}}{{59}}$ , $\cos = \frac{5\sqrt{59}}{{59}}$

$\sin = \frac{\sqrt{59}}{{59}}$ , $\cos = \frac{\sqrt{59}}{{59}}$

$\sin \frac{6}{5}, \cos = \frac{5}{6}$

$\sin = 6\sqrt{59}, \cos \sqrt{59}$

Explanation

We see that the point on the terminal side is (5,6). So we know that with this point a right triangle is formed with a base that is 5 units long, and a leg that is 6 units high. We can use the Pythagorean Theorem to solve for the hypotenuse that is formed by this triangle and this will tell us the distance of the point from the origin.

$a^{2}+b^{2}=c^{2}$

$5^{2}+6^{2}=c^{2}$

$59=c^{2}$

$\sqrt{59}=c$

Let’s solve for sine first. When working with right triangles recall that $\sin = \frac{opposite}{hypotenuse}$ and we are considering the angle formed by the x-axis and the hypotenuse. So the opposite side is the leg that is 6 units high.

$\sin = \frac{opposite}{hypotenuse}$

$\sin = \frac{6}{\sqrt{59}}$

$\sin = \frac{6\sqrt{59}}{{59}}$

We can solve for cosine if we recall that $\cos =\frac{adjacent}{hypotenuse}$ . Our adjacent side would be the base that is 5 units long.

$\cos =\frac{adjacent}{hypotenuse}$

$\cos = \frac{5}{\sqrt{59}}$

$\cos = \frac{5\sqrt{59}}{{59}}$

Find the sine and cosine of the following angle.

Screen shot 2020 08 21 at 4.56.27 pm

$\sin = \frac{6\sqrt{59}}{{59}}$ , $\cos = \frac{5\sqrt{59}}{{59}}$

$\sin = \frac{\sqrt{59}}{{59}}$ , $\cos = \frac{\sqrt{59}}{{59}}$

$\sin \frac{6}{5}, \cos = \frac{5}{6}$

$\sin = 6\sqrt{59}, \cos \sqrt{59}$

Explanation

$a^{2}+b^{2}=c^{2}$

$5^{2}+6^{2}=c^{2}$

$59=c^{2}$

$\sqrt{59}=c$

$\sin = \frac{opposite}{hypotenuse}$

$\sin = \frac{6}{\sqrt{59}}$

$\sin = \frac{6\sqrt{59}}{{59}}$

We can solve for cosine if we recall that $\cos =\frac{adjacent}{hypotenuse}$ . Our adjacent side would be the base that is 5 units long.

$\cos =\frac{adjacent}{hypotenuse}$

$\cos = \frac{5}{\sqrt{59}}$

$\cos = \frac{5\sqrt{59}}{{59}}$

Find the sine value of $\left (1,\sqrt{3} \right )$ if it is a point on the terminal side of an angle in standard position.

$\sqrt{3}/2$

$-\sqrt{3}/2$

Explanation

This is a 30-60-90 triangle. A 30-60-90 triangle will have leg lengths of $\sqrt{3}$ and 1 and a hypotenuse of 2. To find the sin value, you need to divide the opposite leg length with the hypotenuse (opposite/hypotenuse). Thus, giving you $\sqrt{3}/2$ .

Spl 30 60 90

Find the sine value of $\left (1,\sqrt{3} \right )$ if it is a point on the terminal side of an angle in standard position.

$\sqrt{3}/2$

$-\sqrt{3}/2$

Explanation

Spl 30 60 90

The method of solving for trigonometric functions of an angle given a point on its terminal side only works for acute angles.

Which of the following statements best describes the validity of the statement above?

The statement is true

The statement is false

Explanation

Consider the figure below.

We can form a triangle by dropping a line down from the point (-2,3) perpendicular to the

x axis. Now we have right triangle that has a leg that is 3 units high and a base that is 2

units long. Now we can use the Pythagorean Theorem to solve for the hypotenuse.

We can solve for sine and cosine now. Use the fact that . The opposite side of the angle in this case is the leg that is 3 units high.

And we will use the fact that . The adjacent side to the angle is the base that is 2 units long. We do have to note that we are using and not .

So even though our angle was obtuse, we can still use the same method.

The method of solving for trigonometric functions of an angle given a point on its terminal side only works for acute angles.

Which of the following statements best describes the validity of the statement above?

The statement is true

The statement is false

Explanation

Consider the figure below.

We can form a triangle by dropping a line down from the point (-2,3) perpendicular to the

x axis. Now we have right triangle that has a leg that is 3 units high and a base that is 2

units long. Now we can use the Pythagorean Theorem to solve for the hypotenuse.

We can solve for sine and cosine now. Use the fact that . The opposite side of the angle in this case is the leg that is 3 units high.

And we will use the fact that . The adjacent side to the angle is the base that is 2 units long. We do have to note that we are using and not .

So even though our angle was obtuse, we can still use the same method.

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