Question 1

Can the range of arctan also be from 0 to 180? If we divide the circle into quadrants where 0-90 is 1st, 90-180 2nd, 180-270 3rd and 270-360 4th quadrant than the 1st and 3rd have the same tan and the 2nd and 4th have the same tan. As I understand it range is determined so there is only one value for each tan and 1st and 2nd quadrant don't give the same tan.

Accepted Answer

The range of arctan cannot be from 0 to 180 because the 90 degrees angle is undefined and plus as you said the tans will be different in the first and second and in the third and fourth. tan(theta) can have infinite values and that is why several angles can function for many of the values of tan, therefore in order to have an inverse function, you need to restrict the values to the 1st and 4th quadrant in order to avoid 2 angles giving the same tan and so avoid making the function uninvertible. Plus I think you kinda answered your own question. Hopefully, this helps.

Question 2

How do you find the sin(arctan(2)) without using a calculator? 2 is not a value on the unit circle, so I'm not sure how to do this problem by hand. Another problem like this would be cos(arcsin(1/4)). Is there a way to do these types of problems using right triangles? If so, how?
Thank you!

Accepted Answer

sin(arctan 2)
Let's take a look at what this question is really asking for. First, we are taking the arctangent of 2. That is, we are finding some angle 𝜃 such that tan 𝜃 = 2. Then, after finding that angle, we are taking the sine of that angle! In other words, if we have an angle 𝜃 such that tan 𝜃 = 2, we must find sin 𝜃.

This is possible to do with a right triangle. If 𝜃 is an angle in a right triangle such that the opposite side is 2 and the adjacent side is 1, then tan 𝜃 = 2. Then, the sine of this angle is 2/ℎ such that ℎ is the hypotenuse of the triangle. By the Pythagorean theorem, the hypotenuse is √5. Thus, sin 𝜃 = 2/√5 and we have:
sin(arctan 2) = 2/√5
You can use a similar argument for the second question. You could have also solved this using the Pythagorean trig identities (which are basically condensed versions of the process we used above).

Question 3

All I can figure out is what quadrant arcsin (0.98) is in. I don't understand where the 1.37 comes from.

Accepted Answer

1.37 is the angle in radians (in degrees it is approximately 78.52º) in which its sine is equal to 0.98 and, in fact, it is also the first angle that has sine of 0.98 if you follow the trigonometric circle counterclockwise from 0 radians (0º) to 2pi radians (360º). Therefore, it is the principal value of arcsin(0.98), which can also be written as sin^-1(0.98).

Question 4

Hello.

For example, I know how to solve for arcsine(45 degrees), arcsin(3pi/4), and arcsin(square root of 3 / 2). But when it comes to solving for arcsin(0.98), I have no idea how to deal with the number 0.98 and I'm totally confused about how the process works. When I read the explanation down below the exercise 1, it didn't tell me how to get the range of [-pi/2, pi/2]. How would you know the range when you are dealing with the number 0.98? And can somebody explain how the process works to get the answer?

Best regards,
Alex

Accepted Answer

I think you are confusing sine and arcsine. The domain of arcsin(x) is [-1, 1], and neither 45º nor 3π/4 is in that set. However, sin(45º) and sin(3π/4) are both √2/2.

At this level, the only way to compute arcsin(0.98) is with a calculator. In calculus, you'll develop techniques to compute it by hand, but that's beyond the scope of precalculus.

Remember what happens to a functions graph when inverted: it's reflected across the line y=x. If we do this to sin(x), we get a wave going up the y-axis. But this isn't a function. So instead of reflecting the entirety of sin(x), we just reflect the part in [-π/2, π/2]. This smaller piece passes the vertical line test, and it's the largest possible piece that does so.

So because the domain of sin(x) is [-π/2, π/2], this is also the range of arcsin(x).

Question 5

why can I writearcsin(sin (4pi/3)) into arcsin(sin (-pi/3))?

Accepted Answer

Because sin(4π/3)=sin(-π/3). Check them on the unit circle.

Question 6

What do you do when the angle isn't within the domain? Do you have to reflect it?

Accepted Answer

You subtract the amount from the domain and work with what is left.

To use an simple example: imagine you were given a function f(x) = x + 1 and told that x must be more than 0 and less than 10. Now imagine you were given x = 1985 and told to plug it into that function. you would subtract 10 from 1985 until you got a number that fit the domain. In this case that would be five (AKA the remainder after dividing by ten) and you would get f(x) = 5 + 1 = 6

I know that seems really complicated. It comes from a field of mathematics called modular arithmetic which is often used in cryptography (that's the field of codes and ciphers). It isn't as complicated as it sounds but is very hard to explain without diagrams. Khan academy has a section on modular arithmetic under the cryptography section in computer science.

Question 7

Why-90degrees≤arcsin(θ)≤90degrees. Doesn't arcsin(0)equal either 180degrees or 0 degrees?
​​

Accepted Answer

Actually, 0 degrees, 180 degrees, 360 degrees, 540 degrees, and any multiple of 0+/-180*n satisfy the function arcsin(0). However, a function cannot have multiple outputs for a single input. A function must map from one input to a single output. https://en.wikipedia.org/wiki/Function_(mathematics)

In order to satisfy this requirement, mathematicians created the convention to limit the range of the arcsin function to between -90 degrees and 90 degrees. All possible values of the function are contained in that domain (i.e. between -1 and 1), and there are no duplicates or multiples in the range.

Question 8

why are they call arc? I looked it up on wiki and cant find more other source :(

Accepted Answer

I believe it is because the inverse trig functions give the measure of the angle, which if it's in radians and in the unit circle, is also the length of the arc made by that angle.

Radians	Degrees
minus, start fraction, pi, divided by, 2, end fraction, is less than or equal to, \arcsin, left parenthesis, theta, right parenthesis, is less than or equal to, start fraction, pi, divided by, 2, end fraction	minus, 90, degrees, is less than or equal to, \arcsin, left parenthesis, theta, right parenthesis, is less than or equal to, 90, degrees
0, is less than or equal to, \arccos, left parenthesis, theta, right parenthesis, is less than or equal to, pi	0, degrees, is less than or equal to, \arccos, left parenthesis, theta, right parenthesis, is less than or equal to, 180, degrees
minus, start fraction, pi, divided by, 2, end fraction, is less than, \arctan, left parenthesis, theta, right parenthesis, is less than, start fraction, pi, divided by, 2, end fraction	minus, 90, degrees, is less than, \arctan, left parenthesis, theta, right parenthesis, is less than, 90, degrees

Trigonometry

Unit 4: Lesson 1

Inverse trigonometric functions review

What are the inverse trigonometric functions?

Range of the inverse trig functions

Check your understanding

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