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# Intro to arctangent

Sal introduces arctangent, which is the inverse function of tangent, and discusses its principal range. Created by Sal Khan.

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• I don't understand how to FIND the angle which has a given tangent. Around he actually starts to get to work on the problem he made up, but it has a slope of -1 so it's very easy to find the angle. What if you're asked to find the inverse tangent of √3 ?
• Sal silently uses the property that y/x=-1, hence y=-x

Try that with sqrt(3):
y/x=sqrt(3)
y=x*sqrt(3)

Solve Pythagoras but substituting y:
x² + (x*sqrt(3))² = 1
x² + x² * 3 = 1
4x² = 1
x² = 1/4 => x=1/2, y=sqrt(3)/2

You should then RECOGNIZE the "square root 3 over 2" value as a side of a 30-60-90 triangle in a unit circle. Since y>x, x is positive and arctan is also positive, it is a 60 degree or pi/3 angle.
• I have been trying to practice this concept but I find it very hard to solve without the table of angules, especially with arctan exercises, since is not always so easy find a slope of sqrt(3)/3 should we be solving them without any help?
• For 45-90-45 and 30-90-60 triangles, I try to memorize the SIDES of the triangles, not neccesarily the values of the trig ratios. That way I only have to memorize six SIDES, not over 20 ratios. With angles outside of these set triangles, just use a calculator.
• Could someone further illustrate why arctan is restricted to QI and QIV, whereas arccos is restricted to QI and QII? I understand why arcsin is restricted to QI and QIV, and I understand why arccos is restricted to QI and QII. Why is arctan not restricted to QI, QIV and QII? For that matter, why not use all four quadrants for arctan (acknowledging that neither arcsin nor arccos use QIII)?
• Hello, asked the exact same question yesterday, because I couldn t find an explanation. I couldnt stop thinking about it because accepting a convention is no enough, I need to understand the why. And well I figured it out, and it is pretty simple actually.
So think of SOH CAH TOA.
You will notice that both sine and tangent are oppositite(which is the rise, or the y coordinate) , in the numerator, wich makes both of them undefined if the angle is pi/2 or - pi/2 bebause the line would be on the y axis(meaning x=0), and you'd divide it by 0.
The cosine is the one for which you need to put the adjacent (x coordinate) on the numerator, so when there is no angle or the angle is 1pi, the result would also be undefined because the y coordinate would be 0.
Now I call them boundaries.
And I thought of one good tip to memorize which one is what, it that phonetically, "cos" sounds like ksss, or x, so the x axis is the boundary, and in "sine" you really hear the y, so the y axis is the boundary. And we know tan TO(A) is similar to sin SO(H), in the way i told you at the beginning of the message.
I really hope I helped you, I am not used to explain things! But I tried anyway.
• do we always have to convert the answar to radians?
if yes, why? if no, what it better?
• You will not always be required to answer in radians. Whether you answer in radians or degrees depends on what is asked of you. Neither one is better than the other. If you are doing a trig problem, and it was not specified whether to use radians or degrees, use whichever one you are more comfortable with. I personally work the problem out in both degrees and radians to check my answer.
• at "", Why do you include the first quadrant in your restriction of Theta? Why wouldn't you restrict your range of Theta to only the fourth quadrant?
• You need to have all the positive and all the negative possibilities. If you exclude Q1 you can't have any positive answers. Q1 covers the positive angles and Q4 covers the negative angles.
• what does he mean by vanilla tangent?
• The figure of speech "vanilla" refers to a popular flavor of ice cream, the idea being "something is as simple as you can get." So "plain vanilla" implies you're talking about the simplest or most common version of something. Other kinds (and combinations) of ice cream get more complex, sometimes much more complex: https://en.wikipedia.org/wiki/Ice_cream

So comparing something to vanilla ice cream is a simplicity metaphor.
(clarifying edits, typos - I swear I read these things before posting :-) )
• I'm having a lot of trouble with this subject. Could somebody walk me through a detailed explanation of this problem; What is the principal value of sin^-1 (-1/2)?
• here:
look at the question in this way
sin^-1 (-1/2)=what angle of sin?(let that be theta)
now break down the question
sin theta =-1/2....now what angle will be equal to -1/2
which is - 30 degree

therefore ur answer is -30 degree
• At , why isn't the triangle in the 2nd quadrant? Does it make a difference?
• That's just a convention - the principal values of arcsin and arctan are in Q1 & Q4 while the principal values of arccos are in Q1 & Q2.