- Trig angle addition identities
- Using the cosine angle addition identity
- Using the cosine double-angle identity
- Using the trig angle addition identities
- Proof of the sine angle addition identity
- Proof of the cosine angle addition identity
- Proof of the tangent angle sum and difference identities
Sal proves the identity cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y). Created by Sal Khan.
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- Is there a real proof of the angle addition formula involving tangent?(28 votes)
- Once you prove the cosine and sine formulas, you can divide by them and simplify to get the tangent addition formula, as tan(x) is sin(x)/cos(x), so tan (x+y) = sin(x+y) / cos(x+y)(50 votes)
- What if the length of AD is not 1? Then how to prove it?(10 votes)
- If you make the length of AD an arbitrary length, say r, you will just have a bunch of rs floating around until they cancel out and you get the identity we want. This is similar to the reason that we usually use the unit circle to find the trig functions of any angle and not just a generic circle with radius r.(41 votes)
- Something's been bugging me and I know i must be thinking about this the wrong way. Wouldn't the Cos(x+y) always equal 0 in a right triangle? Cause x+y have to equal 90 and the cos of 90 is 0.(9 votes)
- You're right.
But we still use the addition formulas to solve other trigonometric equations where the two angles, x and y, wouldn't add to 90.(4 votes)
- Is it worth remembering this proofs?(2 votes)
- why there is no tangent version of this?(4 votes)
- you can find tangent version:
tan(a+b) = sin (a+b) / cos(a+b)
sin (a+b) = sin(a).cos(b) + sin(b).cos(a)
cos(a+b) = cos(a).cos(b) - sin(a).sin(b)
tan(a+b) = (sin(a).cos(b) + sin(b).cos(a)) / (cos(a).cos(b) - sin(a).sin(b))
now divide numerator and denominator by same factor (cos(a).cos(b))
tan(a+b) = ( tan(a) + tan(b) ) / ( 1 - tan(a).tan(b) )(11 votes)
- If a statement says x is an acute angle, cosx=1/2. Is that enough to determine wether or not sinx is positive or negative?(3 votes)
- According to Princeton University, an acute angle is an angle whose measure is between 0° and 90° (or 0 and π/2 radians). This means that even without knowing that
cos(x) = 1/2we already know that sin(x) is positive since sine of anything in that range will output a value between 0 and 1. Since we also know that
cos(x) = 1/2, we can find that
sin(x) = √̅3/2.
- How does this proof account for values for angle x and y when x or y is greater than or equal to 90? The issue I believe I am observing is that since in the proof, angle x is part of a right triangle, so x has to be less than 90 degrees, and the same can be said for angle y. So, in light of this observation, despite the elegance of this proof compared to others i have seen, isn't this proof flawed, in that it isn't inclusive of angles greater than or equal to 90, thus making it an invalid proof for all real number angle values of x and y?(3 votes)
- This way you, at best substitute one law for another, and that's completely different matter, at worst make unwarranted assumptions about validity of some angel-based arithmetic which you yet has to prove (circular reasoning). @weisbed1 made valid point. Proof supplied in this video proves that this formula works for x,y and x+y acute. Nothing more. I posted outline of simple, elementary general prove of cosine addition formula in tips and thanks.(3 votes)
- The sin addition identity makes sense because we solve for the entire y axis. However, cos solves for only part of the x axis? Why are we solving for segment AF and not for AB?(3 votes)
- On the sine angle identity, you never solved for the entire y axis. Think back to the unit circle, if you have learned it. The angle is x + y, which is equal to A. Remember what sine is. Sine isn't the entire y-axis. Instead, it is the opposite side divided by the hypotenuse. Triangle DAF has angle A, along with a hypotenuse of 1. Since the hypotenuse is 1, you are just dividing the opposite side by 1 to get the sine, which means the the sine is the length of the opposite side in this case. That is why DF is the sine of x + y.
As for this problem, you now have to remember what cosine is. Cosine is the adjacent side divided by the hypotenuse. Again, the hypotenuse of the triangle with x + y is 1. Therefore, you are just dividing the side adjacent to the angle by 1, so the cosine is equal to the adjacent side, which is AF.(1 vote)
- Hi, can you make a video or show how to prove of the tangent identity?(2 votes)
- Let's assume we've already established the addition identities for sine and cosine:
sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
cos(A+B) = cos(A)cos(B) - sin(A)sin(B).
Using the fact that tangent equals sine over cosine, and dividing every term on the top and bottom by cos(A)cos(B) gives us:
tan(A+B) = sin(A+B)/cos(A+B)
= [sin(A)cos(B) + cos(A)sin(B)]/[cos(A)cos(B) - sin(A)sin(B)]
= [sin(A)/cos(A) + sin(B)/cos(B)]/[1 - sin(A)/cos(A) * sin(B)/cos(B)]
= [tan(A) + tan(B)]/[1 - tan(A)tan(B)].(2 votes)
- is there any extra proof to these function(sine cosin )which is more sophisticated(1 vote)
Voiceover: In the last video we proved the angle addition formula for sine. You could imagine in this video I would like to prove the angle addition for cosine, or in particular, that the cosine of X plus Y, of X plus Y, is equal to the cosine of X. Cosine of X, cosine of Y, cosine of Y minus, so if we have a plus here we're going to have a minus here, minus sine of X, sine of X, sine of Y. I'm going to use a very similar technique to the way I proved it for sine. I encourage you to pause the video either now or at any time that you get the inspiration to see if you can do this proof on your own. Just like we thought about it for sine, what is the cosine of X plus Y ^in this diagram right over here? ^Well, X plus Y is this angle right over here. ^If we look at the right triangle ADF, cosine of X plus Y. ^Well, cosine is adjacent over hypotenuse segment AF ^over the hypotenuse, or since the hypotenuse ^is just one, AF divided by one is just going to be AF. ^Cosine of X plus Y is just the length of segment. ^It's just the length of segment AF. ^That right over there is equivalent to this right over here. Let me actually write that down; copy and paste. ^Length of segment AF is cosine of X plus Y. ^Let's think about how we can get that. ^The way I'm going to think about it is, ^given the other right triangles we have in this diagram, ^if we could figure out that or AF. Let me write it this way. This thing, which is the same thing as AF, is equal to. Let me write it this way, it's equal to length of segment AB. It's equal to the length of segment AB, ^which is this entire segment right over here, ^minus the length of segment FB. ^Minus this segment right over here. Minus the length of segment FB. Just from the way our angle addition formula looks for cosine, you might guess what's going to be AB and what's going to be FB. If we can prove that AB is equal to this. If we can prove that FB is equal to this, then we're done. Because we know that cosine of X plus Y, ^which is AF, is equal to AB minus FB. If we can prove this, that it's equal to that minus that. Let's think about what these things actually are. What is AB? ^Let's look at right triangle ACB. We know from the previous video that since triangle ADC has a hypotenuse of one, this length is one, that AC is cosine of X. ^What is AB, well, think about it. ^AB is adjacent to the angle that has measure Y. We could say that the cosine, let me do it down here. We could say, since I've already looked at all of that. ^We could say that the cosine of Y, ^cosine of Y is equal to it's adjacent side. ^The length of it's adjacent side. ^That is segment AB over the hypotenuse ^over cosine of X, cosine of X. ^Or multiply both sides by cosine of X. ^We get that segment AB is equal to cosine of X, ^cosine of X times cosine of Y, cosine of Y, ^which is exactly what we set out to prove. We've just proven that AB is indeed the length of segment AB is indeed equal to cosine X, cosine Y. ^This whole thing is equal to cosine X, cosine Y. ^Now we just have to prove that segment FB is equal to sine X, sine Y. ^This looks like a bit of a strange segment right over here. It's not part of any at least right triangle where we know that I've drawn where we know one of the angles. We can see from this diagram ECBF is a rectangle. We use that fact in the proof for the angle addition formula for sine. We'll also use it now because that tells us that FB is the same as EC, is EC. What is EC going to be equal to? Well, we have this angle Y right over here. What is, let's see, this side is opposite the angle Y. We might want to involve sine. We know that sine of Y, sine of Y, I'm looking right over here, is equal to the length of the opposite side, which is the length of EC, which is the length of EC over the hypotenuse, which is sine of X, sine of X. We figured that out from the last video. If this is X, opposite over hypotenuse is sine of X. Well, the opposite is just one, so the opposite is equal to sine of X. Over here multiply both sides by sine of X, and we get what we were looking for. EC is equal to sine of X, sine of X, times sine of Y, times sine of Y. Once again EC was the exact same thing. Has the same length as segment FB. We have just shown that segment FB is equal to sine of X times sine of Y. This is equal to that right over there. Once again, cosine of X plus Y, which is equal to segment AF, is equal to the length of segment AB minus the length of segment FB, which is equal to, we've proven the length of segment AB is cosine X, cosine Y, minus the length of segment FB, which is sine of X, sine of Y, and we are done.