- 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
Sal is given a drawing with two triangles and some side lengths, and he uses the sine angle addition identity in order to find a missing side length. Created by Sal Khan.
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- In what video does Sal go over the trig identities involved here? I've watched all the videos up to this, but for the life of me can't remember where we learned that sin(alpha + beta) = sin(alpha) * cos(beta) + cos(alpha) * sin(beta)(85 votes)
- Sal says "we know that ..." and tells us the formula for sin(alpha+beta), but in this current trigonometry flow of lessons, this was not covered. It feels like a video is missing in this series, where sal introduces what sin(a+b) is for the first time, or the videos are not in the correct order.(29 votes)
- The proof has snuck ahead to the next section:
- hi sal,how sin (a+b)=sin a cos b+cos a sin b come from? what is to concept to create that formulas?(11 votes)
- Watch this video for the proof of *sin(A+B) = (sin A)(cos B) + (cos B)(sin B)*
It is Khan Academy video. It might help(It's not spam, mark my words).(1 vote)
- If f(x)= (sinx-cosx)^2, the maximum value of f(x) is?(2 votes)
The easiest way to solve this problem is using calculus. Have you studied that yet? If so, we know that we can find local maxima and minima of a function by taking the derivative of it, and solving for x where the derivatives equal 0. So the derivative of (sin x - cos x) ^2 is:
2(sin x - cos x) (cos x + sin x) which equals:
2 (sin x - cos x) (sin x + cos x) which equals:
2 [(sin x)^2 - (cos x)^2] which equals:
2 [(sin x)^2 - ( 1 - sin x^2)] which equals:
2 [ 2(sin x)^2 - 1] which equals:
4 (sin x)^2 - 2 So now we set that equal to 0 and solve for x:
4 (sin x)^2 - 2 = 0 which equals:
4(sin x)^2 = 2 which equals:
(sin x)^2 = 2/4 which equals:
(sin x) ^2 = 1/2 which equals
sin x = +√2/2 or -√2/2.
That means that we have local maxima or minima at values whose sines are √2/2 or -√2/2. That's pi/4 and -pi/4 (and all their multiples, like 3pi/4, -3pi/4, 5pi/4, etc.). But which one is which? Is it a maximum or a minimum?
Well, at pi/4, sin x = √2/2 and cos x = √2/2 so sin x - cos x =0. And 0 squared is just 0.
But at -pi/4, sin x = -√2/2 and cos x = √2/2, so sin x - cos x = -√2, which when squared is 2 (positive 2, because we squared it, so the negative sine went away). So we have a local minimum at pi/4 and a local maximum at -pi/4.
But we also know that we will have other local minimums at all the other multiples of pi/4 where both the sine and cosine have the same sign (they are both positive or they are both negative) and we can figure out that we will have local maxima at all the other multiples of pi/4 where they have opposite signs.
So we have maximas at -pi/4, 3pi/4, 7pi/4, 11pi/4, etc. And at all of those values of x, y = 2. So that's your maximum value! Check out the graph of the function:
- Isn't 0.5 x 0.6 equal to 3? why is it 0.3(0 votes)
- When you multiply a decimal number, you multiply it like normal, then count the decimal places and add it to your result.
0.5 x 0.6 = 5 x 6 = 30 - multiply like normal
0.5 has 1 decimal place, and 0.6 has one decimal place, that makes 2 decimal places.
30, adding 2 decimal places is 0.30 = 0.3
Careful when counting decimal places, for example, 0.50 is the same as 0.5 therefore 0.50 has only 1 decimal place and not 2.(5 votes)
- Are the blue side lengths wrong? Shouldn't they be reversed? The one labeled .6 is clearly longer than the side labeled .8 The answer he presents works out to approx .993 but the hypotenuse is 1 so that side can't be approx 1. If you switch the .6 and .8 of the blue sides then you get the yellow side length to be approx .92 which makes more sense.(2 votes)
- can i practice trigonometry worksheets(1 vote)
- You can find exercises to work with trigonometry right here on Khan Academy! Check out the trig exercises here:
- I apologize if this was asked down the thread, but looking at this problem, my initial instinct was to just break the yellow line into two pieces and use the info given to figure them out independently. But I must be missing something:
I first figured that angle alpha would also be in the top right. I used vertical angles and assume that alpha would need to be the top angle as well. Alpha is 30 degrees based on 30/60/90, so the same is true for that angle. That means that the cos of alpha is equal to .8 over x (which is the top half of the yellow line (which we also know is root 3 over 2 from the bottom triangle). So the top half becomes 4/5 times root 3 over 2 or 4 root 3 over 10.
But here is where I can't make it work - If that top right triangle is also 30/60/90, it means that the short side is .4. That would mean that the other part of that green line should be .2. That .2 then becomes the hypotenuse of the bottom right triangle. And using sine, I should be able to get the second yellow part. Sin of alpha is missing yellow part over .2 which I know is also equal to .5. So I get the second missing yellow part as .1.
My final answer is 1/10 plus 4 root 3 over 10.
Following Sal, I get how he did it, but I can't figure out why my method wouldn't work.(1 vote)
- You're making a couple of mistakes here:
1) You are right about top-right triangle being 30-60-90, but that would mean that it's hypotenuse is
√3/2(cos = adj/hyp ; hyp = adj/cos), or
2) Short side would be equal to half of the hypotenuse, not half of the adjacent side.
16√3/30 / 2 = 8√3/30.
Then the hypotenuse of smaller lower 30-60-90 would be
6/10 - 8√3/30 = (18-8√3)/30, and the length of the opposite side is
(18-8√3)/30 / 2 = (9-4√3)/30. Adding it up to the hypotenuse of the top-right 30-60-90 we get:
(9-4√3)/30 + 16√3/30 = (9+12√3)/30 = 3(3+4√3)/30 =
- What will happen if I moved the yellow vertical line towards the alpha and beta resulting in the reduction of its length? Is your formula still applicable considering the fact that the length of the line segment was reduced?(1 vote)
Voiceover:What I want to do in this video is use all of our powers, all of our knowledge of trig functions and trig identities. In order to figure out, given all the information that we have here, in order to figure out the length of this yellow line, this point, this line or the segment that goes from here to here. I encourage you to pause the video and think about it before I work through it. I'm assuming you've had a go at it and in doing that you might have realized, okay this line that's one of the sides of this right triangle that I have right over here. We're given this alpha and beta but if we consider the combined angle alpha plus beta, then this side right over here, we can just take out our traditional trig functions. Our soh cah toa definition of the basic trig functions and we know that sine is opposite over hypotenuse. If we're considering alpha plus beta, this angle right over here, opposite over hypotenuse that's going to be this length over the hypotenuse which is one. Sine of alpha plus beta is going to be this length right over here. That seems interesting, so let me write that down. Sine of alpha plus beta is essentially what we're looking for. Sine of alpha plus beta is this length right over here. Sine of alpha plus beta it's equal to the opposite side, that over the hypotenuse. Well the hypotenuse is just going to be equal to one so it's equal to this side. Another way of raising the exact same problem that we first tried to tackle is how do we figure out the sine of alpha plus beta? If you're familiar with your trig identities something might be jumping out of you. Hey, we know a different way of expressing sine of alpha plus beta. We know that this thing is the same thing as the sine of alpha times the cosine of beta plus the other way around, the cosine of alpha times the sine of beta. Let me a draw a line here so we don't get confused. If we're trying to figure this out, and we know that this can be re-expressed this way it all boils down to can we figure out what sine of alpha is, cosine of beta, cosine of alpha, and sine of beta. Now when you look at this, you see that you actually can figure those things out. Let's do that. Sine of alpha, I'll write it over here. Sine of alpha is equal to ... This is alpha, sine is opposite over hypotenuse. It's 0.5 over one. This is equal to 0.5, so that is 0.5. Cosine of beta this is beta. Cosine is adjacent over hypotenuse. This is beta, the adjacent side is 0.6 over the hypotenuse of 1. It's 0.6. Cosine of alpha. Adjacent over hypotenuse, it's square root of three over two, over one. That's just square root of three over two. This is just square root of three, over two. Then finally sine of beta. Opposite over hypotenuse is 0.8. This is 0.8. Actually let me write that as, I'm going to write that as four-fifths just so that that's the same thing as 0.8. Just because I think it's going to make it a little bit easier for me to simplify right over here. What is all of these equal to? Well this is going to be equal to 0.5 times 0.6, this part right over here is .3, and square root of three over two, times four-fifths well let's just multiply them. Four divided by two is two. It's two square roots of three over five. This is equal to or so plus two square roots of three over five. This is essentially our answer, I feel a little uncomfortable having it in these two different formats where I have a fraction here and I have a decimal here so let me just write the whole thing as a rational expression. 0.3 is obviously the same thing as three-tenths. That's the same things as three over ten plus, now this if I want to write over ten this the same things as four square roots of three over ten and of course, if we add these two we are going to get three plus four square roots of three, all of that over ten and we are done.