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## 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

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# Finding trig values using angle addition identities

CCSS.Math:

Sal finds the value of sin(7π/12) by rewriting it as sin(π/3+π/4) and then using the sine angle addition formula. Created by Sal Khan.

## Want to join the conversation?

- If the angle cannot be decomposed into pi/4, pi/6 and pi/3 is there a way to find the sine without using a calculator? For example 7pi/11.(22 votes)
- You can use the Taylor Series for sin centered at 0:

x - (x^3)/3! + (x^5)/5! - (x^7)/7! + (x^9)/9! - (x^11)/11!, etc.

Plugging in 7pi/11, you get:

7pi/11 - ((7pi/11)^3)/3! + ((7pi/11)^5)/5!, etc.

It's a lot of work for 7pi/11, but it still works.

Hope this helps!(40 votes)

- So which videos cover the angle addition identities? I'm confused. I never saw any of the videos he references at the beginning of this one that cover the angle addition identities, and can't seem to find them upon searching the site. Furthermore, in the following quiz, there are questions regarding tangent addition identities, which aren't covered in any video I've seen yet on this site, let alone this one. If anyone could point me in the right direction, that would be great.(27 votes)
- I agree. I just had to accept the basic Trig identities that were referenced, as you mentioned, and I came back to the comments in this video because of the tan angle sum/difference identity I also encountered in the practice. I found the tan angle identities in this reference at the end of this section.

https://www.khanacademy.org/math/trigonometry/trig-equations-and-identities/using-trig-identities/a/trig-identity-reference

I haven't looked at the videos between this video and the reference yet but I am expecting this reference to help me through this section.(10 votes)

- In the video, couldn't the magenta line be refereed to as the radius since it's extending from the center. I understand what this video is saying. But I was just curious why the magenta line isn't 1.(0 votes)
- It's extending from the center, but it doesn't reach the circle itself. It's almost the length of the radius, but not quite.(30 votes)

- Edit: I know this question was 3 years ago. At that time I was just relearning math and couldn't see it. Now looking back at this question. It was so simple. All we need to do is factor out √(3) from the numerator and denominator.

------------------------------------------------------------

Original question:

How do you convert (3-√(3)) / (3+√(3)) to (√(3)-1) / (√(3)+1)?

The problem asked to find the tan(15°) using angle addition identities.

My approach was using tan(45°-30°) which led to the answer (3-√(3))/(3+√(3)).

The answer choice to the problem is (√(3)-1) / (√(3)+1) by using the approach tan(60°-45°).

I plugged both answers into the calculator, both gave the exact same value of 0.267949192. Which means they are same answer written in different forms. So how do you algebraically manipulate so that one transform into another?(12 votes)- You factorise the square root of 3 out of the first equation from the numerator and denominator to get root 3 divided by root 3 and then multiplied by the second equation. The root of 3 divided by itself is just 1 and thus you are left with the second equation alone...(7 votes)

- Can this technique be used for every angle?(7 votes)
- Not quite. The angles that they're picking are ones that can be made by adding angles that are easy to remember, namely pi/6, pi/4, pi/3, and pi/2 (30, 45, 60, and 90, respectively) and their multiples. You can use angle addition to quickly find the trig values of, say, 75 degrees, since it's easy to see that 45+30=75.

However, if you are trying to find the trig value of, say, 33 degrees, the angle addition identities won't help you much since you can't add or subtract any of the numbers mentioned above to get to 33. In short, angle addition identities are only good for figuring quick trig values if the angle can be added to (or subtracted to) using the angles mentioned above, or multiples of them.(10 votes)

- I am having problems deciding whether to use addition or subtraction formula with larger angles. For example, if the problem asks for cos(165) why can't I add 135 and 30? The problem suggested 225-60. The hints do not explain how to choose the correct formula. I am very confused with this please help!!(7 votes)
- as long as your sum equals 165, it should work. there are multiple ways to rewrite 165 using the "known angles" (30, 45, 60, 90, 120, 135...), but they all end up the same. if you check 135+30 and 225-60, they are the same thing.(5 votes)

- How do you know what the angle looks like at0:32? I get lost round that part of the video.(3 votes)
- You need to become more familiar with the unit circle and radians.

Radians are actually easier to work with than degrees.

The video mentioned 7π/12, right?

Rather than think of the circles as having 360 degrees, it has 2π radians.

That means π is 180 degrees, and π/2 is 90 degrees.

Now 7/12 is just a bit more than 6/12, and 6/12=1/2, right?

So 7π/12 is just a wee bit more than 6π/12, which is π/2 which is 90 degrees - so 7π/12 is just a wee bit more than 90 degrees.

It might seem complicated at first, but it is way easier, especially when you get into higher math, where degrees are never used.

Try these:

https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/intro-to-radians-trig/v/introduction-to-radians

https://www.mathsisfun.com/geometry/radians.html(9 votes)

- Doesn't a 45 45 90 triangle have legs with a length of 1/sqrt(2)(4 votes)
- Not all 45-45-90 triangles are congruent, so they won't all have the same side lengths. As far as the standard reference triangle, some people think that the sides are 1/sqrt(2) - 1/sqrt(2) - 1, and some people think that it is 1 - 1 - sqrt(2). Those are both similar right triangles with 45 degree acute angles, and the trig and other math will work out on them in exactly the same way.(2 votes)

- Does cos(3pi/4)=-sqrt(2)/2 or +sqrt(2)/2?

Thanks.(3 votes)- Picture the four quadrants of a graph. In quadrant I, both x and y are positive, (+,+). In II, its (-,+), III is (-,-), and in IV, its (+,-).Now you can substitute cosine for x, and sine for y, and the polarity in each quadrant still applies.(3 votes)

- Could this also be answered by using sin(pi/4 + pi/6) since sin(5pi/12) is equivalent to sin(7pi/12)(3 votes)
- Yes, you could do it that way and get the same answer, but either way, you would need to deal with sqrt 3 and sqrt 2 being multiplied and added simultaneously. That is because sin(x) = cos(pi/2 - x), so pi/6 and pi/3 are essentially interchangeable.(3 votes)

## Video transcript

Voiceover:What I want to
attempt to do in this video is figure out what the
sine of seven pi over 12 is without using a calculator. And so let's just
visualize seven pi over 12 in the unit circle. One side of the angle is going
along the positive x-axis and then let's see, if we go straight up, that's pi over two, which is the same thing as six pi over 12, so then we essentially just
have another pi over 12 to get right over there. This is the angle that we're talking about that is seven pi over 12 radians, and the sine of it, by the unit circle definition of sine, it's the y-coordinate of where this ray intersects the unit circle. This is the unit circle, has radius one where it intersects the unit. The y-coordinate is the sine. Another way to think about it, it's the length of this
line right over here. I encourage you to pause
the video right now and try to think about it on your own. See if you can use your
powers of trigonometry to figure out what sine
of seven pi over 12 is or essentially the length
of this magenta line. I'm assuming you've given a go at it, and if you're like me, your first temptation might have been just to focus on this
triangle right over here that I drew for you. The triangle looks like this. It looks like this, where that's what you're
trying to figure out, this length right over here, sine of seven pi over 12. We know the length of
the hypotenuse is one. It's a radius of the unit circle. It's a right triangle right over there. We also know this angle right over here, which is this angle right over here, this gets us six pi over 12, and then we have another pi over 12, so we know that that is pi over 12, not pi over 16. We know that this angle right
over here is pi over 12. Given this information, we can figure out this, or we can at least relate
this side to this other side using a trig function
relative to this angle. This is the adjacent side. The cosine of pi over 12 is going to be this magenta side over one, or you could just say it's
equal to this magenta side. You could say that this
is cosine of pi over 12. We just figured out that
sine of seven pi over 12 is the same thing as cosine of pi over 12, but that still doesn't help me. I don't know offhand what the cosine of pi over 12 radians is without using a calculator. Instead of thinking about it this way, let's see if we can compose this angle or if we can decompose it into some angles for which we do know the sine and cosine. What angles are those? Those are the angles in
special right triangles. For example, we are very familiar
with 30-60-90 triangles. 30-60-90 triangles look
something like this. This is my best attempt
at hand drawing it. Instead of writing 30-degree side, since we're thinking in radians, I'll write that as pi over six radians. The 60-degree side,
I'm going to write that as pi over three radians, and of course, this is the right angle. If the hypotenuse here is one, then the side opposite the 30-degree side, or the pi over six radian side, is going to be half the hypotenuse, which, in this case is 1/2. Then the other side that's
opposite the 60-degree side or the pi over three radian side, is going to be square root of
three times the shorter side. It's going to be square
root of three over two. We've used these types
of triangles in the past to figure out the sine
or cosine of 30 or 60, or in this case, pi over six or pi over three. We know about pi over
six and pi over three. We also know about 45-45-90 triangles. We know that they're
isosceles right triangles. They look like this, my best attempt at drawing it. That one actually doesn't
look that isosceles, so let me make it a little bit more ... I don't know. That looks closer to being
an isosceles right triangle. We know if the length of
the hypotenuse is one, and this comes straight
out of Pythagorean theorem, then the length of each
of the other two sides are going to be square
root of two over two times the hypotenuse, which, in this case, is the
square root of two over two. Instead of describing
these as 45-degree angles, we know that's the same
thing as pi over four, pi over four radians. If you give me pi over six,
pi over three, pi over four, I can use these triangles either using the classic definition, SOHCAHTOA definitions, or I could stick them
on the unit circle here to use the unit circle
definition of trig functions to figure out what the
sine, cosine, or tangent of these angles are. Can I decompose seven pi
over 12 into some combination of pi over sixes, pi over
threes, or pi over fours? Think about that. Let me rewrite pi over six, pi over three, and pi over four with a denominator over 12. Let me write that. Pi over six is equal to two pi over 12, pi over three is equal to four pi over 12, and pi over four is equal
to three pi over 12. Let's see. Two plus four is not seven, two plus three is not seven, but four plus three is seven. So I could use this and this. Four pi over 12 plus three pi over 12 is seven pi over 12. I could rewrite this. This is the same thing as
sine of three pi over 12 plus 4 pi over 12, which, of course, is the same thing, sine of pi over 4, I'll do this in another color, sine of pi over 4 plus ... let me do this ... plus pi over three, Now we can use our angle
addition formula for sine in order to write this
as the sum of products of cosines and sines of these angles. Let's actually do that. This right over here is
going to be equal to, this is going to be equal to the sine, the sine of pi over four times
the cosine of pi over three plus the other way around, cosine of pi over four times
the sine of pi over three, sine of pi over three. Now we just have to
figure out these things, and I've already set up
the triangles to do it. What is sine of pi over four? Sine of pi over four, well, let's think about ... This is pi over four right over here. Sine is opposite over hypotenuse. That's just going to be
square root of two over two. This is square root of two over two, square root of two over two. What is cosine of pi over three? This is a pi over three
radian angle right over here. Cosine is adjacent over a hypotenuse. It's adjacent over a hypotenuse, so this is going to be 1/2. What is cosine of pi over four? Go back to pi over four. It's adjacent over a hypotenuse. It's square root of two over two. It is also square root of two over two, square root of two over two. What's sine of pi over three? Sine is opposite over a hypotenuse, so square root of three over two over one. Square root of three
over two divided by one, which is square root of three over two. Now we just have to simplify
all of this business. This is going to be
equal to the sum of this, or the product, I should say, is just square root of two over four, and then plus the product of these. Let's see. We could write that as
square root of six over four, square root of six over four, or we could just rewrite
this whole thing as, and we deserve a little bit
of a drum roll at this point, this is equivalent to, let me just scroll over
to the right a little bit. This is equivalent to square root of two plus square root of six, all of that over four. That's what sine of seven pi over 12 is, or cosine of pi over 12, what that is equal to.