Trigonometric Equation

harpazo

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Given cos x = k, where k > 0, use the reference-angle concept and a calculator to find another root of the equation (within the interval 0 less than or to x less than or equal to 2pi). Round answer to 4 decimal places. How is this done? What is the reference-angle concept?

Note: cos x = 0.9

This tells me that k = 0.9.
 
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MarkFL

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I would observe that the angle \(x\) is symmetric about 0. Specifically:

\(\displaystyle x=0\pm\arccos(k)+2\pi n=2\pi n\pm\arccos(k)\) where \(n\in\mathbb{Z}\)

On the stated interval, this gives us:

\(\displaystyle x=\arccos(k),\,2\pi-\arccos(k)\)
 

HallsofIvy

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I can't imagine a textbook having a problem asking for a "reference angle" if it hasn't previously defined "reference angle"! The trig functions are originally defined in a right triangle, so for angles between 0 and \(\displaystyle \pi/2\) radians (90 degrees) but then extended to any angle. The "reference angle" when calculating a trig function of an angle is the angle between 0 and \(\displaystyle \pi/2\) radians that gives the same value to that trig function.
 

harpazo

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I can't imagine a textbook having a problem asking for a "reference angle" if it hasn't previously defined "reference angle"! The trig functions are originally defined in a right triangle, so for angles between 0 and \(\displaystyle \pi/2\) radians (90 degrees) but then extended to any angle. The "reference angle" when calculating a trig function of an angle is the angle between 0 and \(\displaystyle \pi/2\) radians that gives the same value to that trig function.
By "an angle between 0 and pi/2 radians" you mean an angle in quadrant 1 of the unit circle forming 90 degrees. Yes?
 

harpazo

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I would observe that the angle \(x\) is symmetric about 0. Specifically:

\(\displaystyle x=0\pm\arccos(k)+2\pi n=2\pi n\pm\arccos(k)\) where \(n\in\mathbb{Z}\)

On the stated interval, this gives us:

\(\displaystyle x=\arccos(k),\,2\pi-\arccos(k)\)
1. What do you mean the angle x is symmetric about 0? Do you mean symmetric about the origin?

2. I do not understand your second line starting with x = 0. Can you explain this math statement in words?

3. I do not understand your the for x in your third line involving arc cosine.
 

MarkFL

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1. What do you mean the angle x is symmetric about 0? Do you mean symmetric about the origin?

2. I do not understand your second line starting with x = 0. Can you explain this math statement in words?

3. I do not understand your the for x in your third line involving arc cosine.
This is what I see in my mind when presented with:

\(\displaystyle \cos(x)=k\) where \(0<k\)


Take a look at that and see if this answers your questions about my initial post...
 

harpazo

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Otis

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Given an angle θ, the reference angle is the smallest angle between the terminal ray of angle θ and the x-axis.

This chart shows reference angles, for some example angles θ (one image for each quadrant).

refAng.JPG

😎
 

harpazo

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Given an angle θ, the reference angle is the smallest angle between the terminal ray of angle θ and the x-axis.

This chart shows reference angles, for some example angles θ (one image for each quadrant).

View attachment 19052

😎
Cool pictures. Using algebra, how do I find the reference angle for each of the following questions?

A. 240°

B. 4pi/3
 

Otis

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Cool pictures. Using algebra, how do I find the reference angle for … 240°
In what quadrant does the terminal ray of 240º lie? That's the third quadrant.

Look at the previous image, for QIII. Pretend that θ is 240º. The reference angle (shown in red) rotates from 180º to 240º. How many degrees is that?

240 - 180 = 60

The reference angle for 240º is 60º.

We may take the same approach, to find the reference angle for 4pi/3. In what quadrant does the terminal ray of 4pi/3 lie?

😎
 

harpazo

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In what quadrant does the terminal ray of 240º lie? That's the third quadrant.

Look at the previous image, for QIII. Pretend that θ is 240º. The reference angle (shown in red) rotates from 180º to 240º. How many degrees is that?

240 - 180 = 60

The reference angle for 240º is 60º.

We may take the same approach, to find the reference angle for 4pi/3. In what quadrant does the terminal ray of 4pi/3 lie?

😎
The terminal ray of 4pi/3 lies in quadrant 3.

Let R = reference angle

R = given angle - pi

R = (4pi/2) - (pi/1)

R = pi/3 radians

Yes?
 

Otis

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The terminal ray of 4pi/3 lies in quadrant 3.
Yes, that's correct. 3pi/3 is half a revolution, so 4pi/3 is pi/3 more than half a revolution, and that puts the terminal ray in QIII. (That pi/3 rotation beyond half a revolution is the reference angle; review the image.)

R = given angle - pi
R = (4pi/2) - (pi/1)
R = pi/3 radian
Your setup is correct, for an angle in QIII (angle - pi). Yes, the reference angle is pi/3.

Minor typo: the given angle is 4pi/3, not 4pi/2.
 

harpazo

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Yes, that's correct. 3pi/3 is half a revolution, so 4pi/3 is pi/3 more than half a revolution, and that puts the terminal ray in QIII. (That pi/3 rotation beyond half a revolution is the reference angle; review the image.)


Your setup is correct, for an angle in QIII (angle - pi). Yes, the reference angle is pi/3.

Minor typo: the given angle is 4pi/3, not 4pi/2.
Yes, I meant type 4pi/3 not 4pi/2. Typo at my end.
 

Otis

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… use the reference-angle concept and a calculator to [solve] the equation [within 0<θ<2pi]. Round answer to 4 decimal places. How is this done? What is the reference-angle concept?

cos(θ) = 0.9
Now that you know what a reference angle is, we can do this. Note: I changed the angle variable from x to θ because I want to talk about x as a coordinate, instead. I'm using function notation cos(), also.

As a terminal ray rotates through one complete revolution 0 < θ < 2pi, it will designate two angles whose cosine is 0.9 and two angles whose cosine is -0.9. Remember, the number cos(θ) is the x-coordinate of the point where ray intersects the unit circle. There are two points on the circle whose x-coordinate is 0.9 and two points whose x-coordinate is -0.9.

When cosine is positive (like 0.9), then the terminal ray must be in QI or QIV because those are the two quadrants where x-coordinates are positive.

When cosine is negative (like -0.9), then the terminal ray lies in QII or QIII because x-coordinates are negative there.

So, we need to find the angles in QI and QIV where cos(θ)=0.9. The reference angle is the angle in QI -- a calculator's inverse cosine function gives us that angle.

arccos(0.9) ≈ 0.45103 (that's about 25.8º)

Now, what is θ in QIV? Can you use the reference angle, to determine it? Review the prior image, if you need to, to see the reference angle drawn (in red) for an angle in QIV.

To calculate θ, what gets subtracted from what? If you do the subtraction, round the result to four places.

😎
 

topsquark

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Cool pictures. Using algebra, how do I find the reference angle for each of the following questions?

A. 240°

B. 4pi/3
A slight correction here. \(\displaystyle 4 \pi / 3\) is an angle so you need to designate it as such by writing \(\displaystyle 4 \pi / 3\) rad.

I realize that radians is a rather weird unit, being that it is actually unitless, but we need the unit to indicate that it is an angle, not just a number.

It's just a good practice to get into. (Even if there are people out there that would disagree.)

-Dan
 

Dr.Peterson

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A slight correction here. \(\displaystyle 4 \pi / 3\) is an angle so you need to designate it as such by writing \(\displaystyle 4 \pi / 3\) rad.

I realize that radians is a rather weird unit, being that it is actually unitless, but we need the unit to indicate that it is an angle, not just a number.

It's just a good practice to get into. (Even if there are people out there that would disagree.)

-Dan
I disagree. Many authors define the trig functions as functions of a real number, not of an angle; and in such a book, when angles are in view and there is no explicit unit, that implies radians. This is standard in pure math.

In applications, such as physics, consistent use of units is more standard, and also wise. But if you are working in a math textbook using the conventions I refer to, it would be wrong to say they are wrong. What you are expressing here is not a "correction", but a preference (though it is one that is justified in the right context).
 

harpazo

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Now that you know what a reference angle is, we can do this. Note: I changed the angle variable from x to θ because I want to talk about x as a coordinate, instead. I'm using function notation cos(), also.

As a terminal ray rotates through one complete revolution 0 < θ < 2pi, it will designate two angles whose cosine is 0.9 and two angles whose cosine is -0.9. Remember, the number cos(θ) is the x-coordinate of the point where ray intersects the unit circle. There are two points on the circle whose x-coordinate is 0.9 and two points whose x-coordinate is -0.9.

When cosine is positive (like 0.9), then the terminal ray must be in QI or QIV because those are the two quadrants where x-coordinates are positive.

When cosine is negative (like -0.9), then the terminal ray lies in QII or QIII because x-coordinates are negative there.

So, we need to find the angles in QI and QIV where cos(θ)=0.9. The reference angle is the angle in QI -- a calculator's inverse cosine function gives us that angle.

arccos(0.9) ≈ 0.45103 (that's about 25.8º)

Now, what is θ in QIV? Can you use the reference angle, to determine it? Review the prior image, if you need to, to see the reference angle drawn (in red) for an angle in QIV.

To calculate θ, what gets subtracted from what? If you do the subtraction, round the result to four places.

😎

To find theta, I would use θ = 2pi - 0.9.

Yes?

If not, can you set it up for me?
 

MarkFL

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To find theta, I would use θ = 2pi - 0.9.

Yes?

If not, can you set it up for me?
No, as per my first post the two angles would be (substituting \(k=0.9\)):

\(\displaystyle \theta=\arccos(0.9),\,2\pi-\arccos(0.9)\)
 

harpazo

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No, as per my first post the two angles would be (substituting \(k=0.9\)):

\(\displaystyle \theta=\arccos(0.9),\,2\pi-\arccos(0.9)\)
My first choice was to subtract cos^(-1) (0.9) from 2pi but I then started to doubt.
If theta is the reference angle, we can say that theta = 2pi - cos^(-1) (0.9).
 

Otis

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To find theta, I would use θ = 2pi - 0.9.
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)

… If theta is the reference angle, we can say that theta = 2pi - cos-1(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 2* 3.14159 - 0.45103 = 5.832155307......................................edited

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

😎
 
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