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## Trigonometry

### Course: Trigonometry>Unit 4

Lesson 5: Using trigonometric identities

# Using trig angle addition identities: finding side lengths

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.

## Want to join the conversation?

• 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) • 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. • hi sal,how sin (a+b)=sin a cos b+cos a sin b come from? what is to concept to create that formulas? • Broooo another bangerrrrrr • If f(x)= (sinx-cosx)^2, the maximum value of f(x) is? • Sumeet,

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 • 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.
• 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. • can i practice trigonometry worksheets
(1 vote) • 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 `8/10` divided by `√3/2` (cos = adj/hyp ; hyp = adj/cos), or `16√3/30`.

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 =`
`= (3+4√3)/10`

ta-da 