Trigonometric Equation

[harpazo]

Not quite because 0.9 is not an angle; it's a trig ratio (cosine output). To get angles, we need to subtract angles. For the angle θ, we need to subtract the angle arccos(0.9) from the angle 2pi:

2pi - arccos(0.9)


Yes, that subtraction is correct, but let's call the reference angle θ' (theta-prime) because we've been using θ to mean the QIV angle.

θ' = arccos(0.9) ≈ 0.45103

θ = 2pi - θ'

That's about 3.14159 - 0.45103 = 2.69056

θ ≈ 2.6906

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This is exactly what I got after playing with the problem on paper.

 

[Otis]

… 2* 3.14159 - 0.45103 = 5.832155307......................................edited

θ ≈ 2.6906 5.832155307......................................edited

To the moderator who fixed my math goof, I say 'thanks'.

?

 

[Otis]

3.14159 - 0.45103 = 2.69056
θ ≈ 2.6906

This is exactly what I got after playing with the problem on paper.

Oops -- I made a mistake, harpazo (subtracting from pi, instead of from 2pi). Did you do "exactly" the same?

(A moderator has since corrected that math goof, in post #20, but the final answer is not rounded to four places.)

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Are you game for some more reference-angle practice?

| sin(x) | ≈ 0.342020

Find all solutions for 0º < x < 360º (round each to the nearest degree).

$\;$

 

[harpazo]

Oops -- I made a mistake, harpazo (subtracting from pi, instead of from 2pi). Did you do "exactly" the same?

(A moderator has since corrected that math goof, in post #20, but the final answer is not rounded to four places.)

---

Are you game for some more reference-angle practice?

| sin(x) | ≈ 0.342020

Find all solutions for 0º < x < 360º (round each to the nearest degree).

$\;$

I subtracted from 2pi on paper.

Oops -- I made a mistake, harpazo (subtracting from pi, instead of from 2pi). Did you do "exactly" the same?

(A moderator has since corrected that math goof, in post #20, but the final answer is not rounded to four places.)

---

Are you game for some more reference-angle practice?

| sin(x) | ≈ 0.342020

Find all solutions for 0º < x < 360º (round each to the nearest degree).

$\;$

Set Up:

Let R = reference angle

R = 2pi - sin^(-1) (0.342020)

Yes?

 

[Otis]

I subtracted from 2pi on paper …

In that case, I'm not sure why you said you also got 2.6906 (the wrong answer).

?

 

[Otis]

| sin(x) | ≈ 0.342020
Find all solutions for 0º < x < 360º (round each to the nearest degree).

… Let R = reference angle
R = 2pi - sin^(-1) (0.342020)

In this exercise, we're working with degree measure.

A reference angle is always between 0º and 90º. Look at the image, again. In each quadrant, the reference angle is shown in red. Can you see they are always smaller than a right-angle?

Your calculation for R (above) looks like a calculation for the angle in Q4 (subtracting an angle from 360º).

Let's start by finding the reference angle. What angle in QI is a solution? (That angle is the reference angle.)

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[harpazo]

In that case, I'm not sure why you said you also got 2.6906 (the wrong answer).

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I misspoke.

 

[harpazo]

In this exercise, we're working with degree measure.

A reference angle is always between 0º and 90º. Look at the image, again. In each quadrant, the reference angle is shown in red. Can you see they are always smaller than a right-angle?

Your calculation for R (above) looks like a calculation for the angle in Q4 (subtracting an angle from 360º).

Let's start by finding the reference angle. What angle in QI is a solution? (That angle is the reference angle.)

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The reference angle in Q1 is the same as the given angle.

Let R = reference angle

R = 90° - sin^(-1) (given angle)

Is this the set up, Otis?

 

[Otis]

The reference angle in Q1 is the same as the given angle.

I didn't provide any angles, harpazo. What were you looking at, when you thought you saw a given angle?

90° - sin^(-1) (given angle)

That expression gives what we call the complementary angle. The reference angle is something else.

| sin(x) | ≈ 0.342020

That equation says:

sin(x) ≈ 0.342020 $\;$ OR $\;$ sin(x) ≈ -0.342020

sin(x) is the y-coordinate of points on the unit circle.

There are two points on the unit circle whose y-coordinate is 0.342020 (one is in QI, and the other is in QII).

There are two points on the unit circle whose y-coordinate is -0.342020 (one is in QIII, and the other is in QIV).

For each of those four points, we can draw a ray from the Origin through the point. Each of those rays is the terminal ray of an angle in standard position.

The inverse sine function gives us the angle in QI. We call that the reference angle. Look at the image, again. Can you see that the QI angle and the reference angle are the same angle? (That's not true, in the other three quadrants.)

We use the reference angle, to determine angles in other quadrants whose sine is also 0.342020 or -0.342020.

In other words:

QI solution (the reference angle) = arcsin(0.342020)

QII solution = 180º - arcsin(0.342020)

QIII solution = 180º + arcsin(0.342020)

QIV solution = 360º - arcsin(0.342020)

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[harpazo]

I didn't provide any angles, harpazo. What were you looking at, when you thought you saw a given angle?


That expression gives what we call the complementary angle. The reference angle is something else.

| sin(x) | ≈ 0.342020

That equation says:

sin(x) ≈ 0.342020 $\;$ OR $\;$ sin(x) ≈ -0.342020

sin(x) is the y-coordinate of points on the unit circle.

There are two points on the unit circle whose y-coordinate is 0.342020 (one is in QI, and the other is in QII).

There are two points on the unit circle whose y-coordinate is -0.342020 (one is in QIII, and the other is in QIV).

For each of those four points, we can draw a ray from the Origin through the point. Each of those rays is the terminal ray of an angle in standard position.

The inverse sine function gives us the angle in QI. We call that the reference angle. Look at the image, again. Can you see that the QI angle and the reference angle are the same angle? (That's not true, in the other three quadrants.)

We use the reference angle, to determine angles in other quadrants whose sine is also 0.342020 or -0.342020.

In other words:

QI solution (the reference angle) = arcsin(0.342020)

QII solution = 180º - arcsin(0.342020)

QIII solution = 180º + arcsin(0.342020)

QIV solution = 360º - arcsin(0.342020)

?

1. Thank you for this great lesson on reference angle here.

2. You said:

"I didn't provide any angles, harpazo. What were you looking at, when you thought you saw a given angle?"

I was looking at the argument of sine. I completely overlooked the fact that the given question relates to the absolute value of the sine function. I was not paying attention.

 

[Otis]

I was looking at the argument of sine …

Ah, I understand now, but that given argument is only a variable symbol.

The given expression sin(x) shows that symbol x has been chosen to represent possible angles. Therefore, we don't call symbol x a "given angle".

When we say "given angle", we're referring to a specific measurement (number of degrees or radians) provided by an exercise statement.

Can you finish the practice exercise?

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[harpazo]

Ah, I understand now, but that given argument is only a variable symbol.

The given expression sin(x) shows that symbol x has been chosen to represent possible angles. Therefore, we don't call symbol x a "given angle".

When we say "given angle", we're referring to a specific measurement (number of degrees or radians) provided by an exercise statement.

Can you finish the practice exercise?

?

Ok. I get it now in terms of the argument of trig functions. Do you want me to complete the following?

You stated:

QI solution (the reference angle) = arcsin(0.342020)

QII solution = 180º - arcsin(0.342020)

QIII solution = 180º + arcsin(0.342020)

QIV solution = 360º - arcsin(0.342020)

Do you want me to calculate the above and show my work per quadrant?

 

[Otis]

… Do you want me to calculate the above and show my work per quadrant?

The exercise asks for four solutions (rounded to the nearest degree). Those are what I'd like to see. You don't need to show work, as I gave you the setups. (You seemed to be having difficulty understanding how to form those expressions, from looking at the reference-angle image).

?

 

[harpazo]

The exercise asks for four solutions (rounded to the nearest degree). Those are what I'd like to see. You don't need to show work, as I gave you the setups. (You seemed to be having difficulty understanding how to form those expressions, from looking at the reference-angle image).

?

QI solution (the reference angle) = arcsin(0.342020)

Answer: 0°

QII solution = 180º - arcsin(0.342020)

Answer: 179°

QIII solution = 180º + arcsin(0.342020)

Answer: 180°

QIV solution = 360º - arcsin(0.342020)

Answer: 360°

 

[Otis]

QI solution (the reference angle) = arcsin(0.342020)

Answer: 0°

Hi. Maybe I misspoke; we may need to see your steps. But first, please tell us the calculator used to evaluate arcsin(0.342020). Was it set to degree-mode?

Here's a tip. We know the Q1 angle cannot be 0° because sin(Q1 angle) is 0.342020. In other words, the y-coordinate of the point where Q1's terminal ray intersects the unit circle is 0.342020.

But, the y-coordinate with 0° is 0. The terminal ray of 0° lies on the horizontal axis, right? (The ray has not rotated at all.)

We need the ray to rotate counterclockwise (from its initial position on the positive, horizontal axis) until the intersection point traveling along the unit circle reaches a height of 0.342020 units above the axis. At that point, the angle is a solution to sin(angle)=0.342020. That's the Q1 angle. The arcsine function outputs the Q1 angle, when the input is the y-coordinate 0.342020 (and the calculator is set to the correct mode).

?

 

[harpazo]

Hi. Maybe I misspoke; we may need to see your steps. But first, please tell us the calculator used to evaluate arcsin(0.342020). Was it set to degree-mode?

Here's a tip. We know the Q1 angle cannot be 0° because sin(Q1 angle) is 0.342020. In other words, the y-coordinate of the point where Q1's terminal ray intersects the unit circle is 0.342020.

But, the y-coordinate with 0° is 0. The terminal ray of 0° lies on the horizontal axis, right? (The ray has not rotated at all.)

We need the ray to rotate counterclockwise (from its initial position on the positive, horizontal axis) until the intersection point traveling along the unit circle reaches a height of 0.342020 units above the axis. At that point, the angle is a solution to sin(angle)=0.342020. That's the Q1 angle. The arcsine function outputs the Q1 angle, when the input is the y-coordinate 0.342020 (and the calculator is set to the correct mode).

?

My answers came from Wolfram. I do not have a scientific calculator, Otis. I know the difference between degree mode and radian mode.

 

[harpazo]

The exercise asks for four solutions (rounded to the nearest degree). Those are what I'd like to see. You don't need to show work, as I gave you the setups. (You seemed to be having difficulty understanding how to form those expressions, from looking at the reference-angle image).

?

Answers:

Inverse sine of 0.342020 ≈ 20°...thus ≈ 20° for quadrant 1.

Quadrants 2 to 4:

200° , 160° , 340°

 

[Dr.Peterson]

QI solution (the reference angle) = arcsin(0.342020)
Answer: 0°

QII solution = 180º - arcsin(0.342020)
Answer: 179°

QIII solution = 180º + arcsin(0.342020)
Answer: 180°

QIV solution = 360º - arcsin(0.342020)
Answer: 360°

My answers came from Wolfram. I do not have a scientific calculator, Otis. I know the difference between degree mode and radian mode.

Interesting. I presume you are saying that you did this: https://www.wolframalpha.com/input/?i=180º+-+arcsin(0.342020)

What this tells us is that Wolfram Alpha doesn't understand the concept of degree mode: it understood that 180 was in degrees, but not that you therefore wanted arcsin(0.342020) in degrees, even though it would make no sense to subtract radians from degrees. I would have expected better!

This, of course, is the trouble with depending on technology; the smarter it is, the more you have to hope it is thinking the same way you are.

What you might want to do (if you don't have a smart phone that can run calculator apps, and don't want to spend the $10 to $20 for a basic scientific calculator -- for which I don't blame you), is to find a less smart calculator site that you can depend on to do just what you tell it to do, and no more. As great as WA is for some purposes, it isn't the best tool for everything. Google "online scientific calculator" to find several.

 

[harpazo]

Interesting. I presume you are saying that you did this: https://www.wolframalpha.com/input/?i=180º+-+arcsin(0.342020)

What this tells us is that Wolfram Alpha doesn't understand the concept of degree mode: it understood that 180 was in degrees, but not that you therefore wanted arcsin(0.342020) in degrees, even though it would make no sense to subtract radians from degrees. I would have expected better!

This, of course, is the trouble with depending on technology; the smarter it is, the more you have to hope it is thinking the same way you are.

What you might want to do (if you don't have a smart phone that can run calculator apps, and don't want to spend the $10 to $20 for a basic scientific calculator -- for which I don't blame you), is to find a less smart calculator site that you can depend on to do just what you tell it to do, and no more. As great as WA is for some purposes, it isn't the best tool for everything. Google "online scientific calculator" to find several.

1. Thank you for your reply.

2. I was place on furlough WITHOUT PAY by the employer on May 16. Currently living off my savings until unemployment benefits kick in, if this ever happens. I cannot purchase a calculator now.

3. Reading 2 above indicates that money is tight, really tight. Holding on by the skin of my teeth just like millions across the planet due to COVID-19. Waiting for Governor Cuomo to unlock NYC so that I can get back to work.

4. I will search for a FREE online calculator.

5. I will do MY VERY BEST MOVING FORWARD to show math work to new threads even if the work makes no sense.

 

[Otis]

… 20° for quadrant 1.
Quadrants 2 to 4:
200° , 160° , 340°

Correct!,

Tip: In future situations when you have a calculator that reports only radians, (if needed) you can convert to degrees -- multiply by 180/pi. For example:

$$\frac{0.3491 \cancel{rad}}{1} × \frac{180^{\circ}}{\pi \cancel{rad}} = \frac{0.3491×180^{\circ}}{3.1416} ≈ 20^{\circ}$$

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