Main content

## Precalculus

### Course: Precalculus > Unit 2

Lesson 10: Using trigonometric identities- Finding trig values using angle addition identities
- Using the tangent angle addition identity
- Find trig values using angle addition identities
- Using trig angle addition identities: finding side lengths
- Using trig angle addition identities: manipulating expressions
- Using trigonometric identities
- Trig identity reference
- Trigonometry: FAQ

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Using the tangent angle addition identity

CCSS.Math:

Find the tangent of 13pi/12 without a calculator using the tangent angle addition identity. Created by Sal Khan.

## Want to join the conversation?

- In1:16, Sal said we've proven tangent identities in another video. Where can I find that video?

I was trying to search the tangent identities but couldn't find a video with the sum and difference identities of the tangent.(3 votes) - Why did Sal rationalize the variables last? I mean, I can see why that was a good idea since it was faster than just rationalizing them from the start, but is there a rule on when to rationalize or is it just common sense on when it's a good idea to rationalize now than later?

Edit: I notice he didn't have to rationalize this way.(2 votes)- It's always easier to work with smaller numbers, so personally I lean to rationalizing denominators afterwards unless I see some obvious benefit of doing it beforehand.(2 votes)

- I have been using ChatGPT for useful discussions. But remember it is not infallible. It makes mistakes that it owns up to when you point it out(1 vote)
- When Sal says the slope of the tangent on the unit circle is just the radius for 5pi/4 = 1, why is it different for pi/6?(1 vote)
- Since tangent is opposite divided by adjacent, the values in the 45-45-90 triangle are both the same so it'll just be 1.(1 vote)

- How would you solve an equation such as (cos(x))/(1+csc(x))*(1-csc(x))/(1-csc(x))?

I end up always getting these wrong, and they always answer (such as with this one) with something like tan(x)-tan(x)sin(x). why and how did they get to that answer??(1 vote) - why are the practice problems so much harder than what we learn in the video?

i feel like there are not enough videos here to go ahead to the parctice problem because he does not show how to solve so many of them(0 votes)- You could read the article at the end of this section, this covers equations ranging from the
*inverse*of sin to the*reciprocal*of tangent and the equations to solve the problems.(1 vote)

## Video transcript

- [Instructor] In this video,
we're going to try to compute what tangent of 13 pi over 12
is without using a calculator. But I will give you a few hints. First of all, you can rewrite
tangent of 13 pi over 12 as tangent of, instead of 13 pi over
12, we can express that in terms of angles where we
might be able to figure out the tangents just based
on other things we know about the unit circle. 13 pi over 12 is the same
thing as 15 pi over 12 minus two pi over 12, which is the same thing as the tangent of five pi over four minus pi over six, or we could even view it as
plus negative pi over six. So that's my hint right over there. So pause this video and
see if you can keep going with this train of reasoning to evaluate what tangent
of 13 pi over 12 is without using a calculator. All right, now let's
keep on going together. Well, we already know what the tangent of the sum of two angles are. We've proven that in another video. We know that this is going to be equal to the tangent of the
first of these angles, five pi over four, plus the tangent of the second angle, tangent of negative pi over six, all of that is going to be over one minus tangent of the first angle, five pi over four, times the tangent of the second angle, negative pi over six. And so now we can break
out our unit circles to figure out what these things are. So I have pre-put some unit circles here. And so let's first think about what five pi over four looks like. Pi over four, you might already associate
it with 45 degrees. That's pi over four right over there. Two pi over four would get you here. Three pi over fours would get you there. Four pi over four, which
the same thing as pi, gets us over there. Five pi over four would
get us right about there. Now, you might already recognize
the tangent of an angle as the slope of the radius, and so you might already be able to intuit that the tangent here is going to be one, but we can also break out
our knowledge of triangles in the unit circle to figure this out if you didn't realize that. So what we need to do is
figure out the coordinates of that point right over there, and to help us do that, we can set up a little
bit of a right triangle, which you might immediately recognize is a 45-45-90 triangle. How do I know that? Well five pi over four, remember we go four pi
over four to get here, then we have one more pi
over four to go down here. So this angle right over
here is pi over four or you could view it as 45 degrees. And of course, if that's
45 degrees and that's 90, then this has to be 45 degrees 'cause they all add up to 180 degrees. And we know a triangle like
this by the Pythagorean theorem, if our hypothesis is one, each of the other two sides is square root of two over
two times the hypotenuse. So this is square root of two over two and then this is square
root of two over two. Now, if we think about the coordinates, our X coordinate is square
root of two over two in the negative direction. So our X coordinate is negative
square root of two over two and our Y coordinate is square root of two over two going down, so that's also negative
square root of two over two. And the tangent is just the Y coordinate over
the X coordinate here. So the tangent is just going to be negative square root of two over two over negative square root of two over two, which is once again, one,
which was our intuition. So we can write the tangent
of five pi over four is equal to one. And then what about negative pi over six? Well, negative pi over six, you might recognize pi over
six as being a 30 degree angle. Pi is 180 degrees, so
divided by six is 30 degrees, and negative pi over six would be going 30 degrees
below the positive x-axis. So it would look just like that. And as we said, this angle
right over here is pi over six, which you can also view
as a 30 degree angle. If we were to drop a
perpendicular right over here, you might immediately recognize this as a 30-60-90 triangle. We know if the hypotenuse
is of length one, the side opposite the 30 degree side is 1/2 the hypotenuse, and then the longer non-hypotenuse side is going to be square root of
three times the shorter side, so square root of three over two. And so our coordinate's right over here, we are moving square
root of three over two in the positive X direction, square root of three over two, and then we're going negative
1/2 in the Y direction, so we put negative 1/2 right over there. And so now we know that the
tangent of negative pi over six is going to be equal to negative 1/2 over square root of three over two, which is the same thing
as negative 1/2 times, let me write it this way, negative 1/2 times two over the square root of three, which is equal to negative one over the square root of three. This right over here is one. We saw that there. This right over here is one. And then this right over here is negative one over the
square root of three. And then this is negative one
over the square root of three. And so I can rewrite this
entire expression as being equal to one minus one over the square to three, all of that over one, and then I have a negative
here and a negative here, so then that becomes, a
negative times a negative, so positive one plus one over
the square root of three. And if we multiply both the
numerator and the denominator by square root of three, what we're going to get in the numerator is square root of three minus one and then our denominator is going to be square root of three plus one. And we are done. That's tangent of 13 pi over
12 without using a calculator.