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## Trigonometry

### Course: Trigonometry>Unit 4

Lesson 6: Challenging trigonometry problems

# Trig challenge problem: system of equations

Sal solves a very complicated algebraic trig problem that appeared as problem 55 in the 2010 IIT JEE Paper I exam. Created by Sal Khan.

## Video transcript

The number of all possible values of theta, where theta is between zero and pi for which the system of equations y plus z cosine of 3 theta is equal x y z sin 3 theta x sin of 3 theta is equal to 2 cos 3 theta over y plus 2 sin 3 theta over z xyz sin 3 theta is equal y plus two z cos of 3 theta plus y sin 3 theta, has a solution x naught y naught z naught with y naught and z naught not equal to zero is, which means either of these cannot equal zero So this is a pretty daunting thing, I just like to simplify this as much as possible. I want to do two things. All of the trigonometric functions here they're all - they're taking sin of, the cosine of three theta and have theta between zero and pi I'll do a little substitution. this is just from my brain,cause it simplifies my logic a little bit let's make a substitution that u is equal to - is equal to three theta and then the u - when theta is zero, u is going to be - so u is still going to be greater than zero and when theta is pi, u is going to - have to be less than three pi And so - let's find the number of Us between zero and three pi where y plus z cosine of u is equal to xyz sine of u so on and so forth. so that - Let's do that substitution and I'm going to rearrange these so that they start at least look a little familiar and see if we can somehow start manipulating with the these equations and actually try to cancel out terms so this first one up here just to get in a term ways i can recognize. let me distribute the cosine of three theta or which will now call the cosine of u. So let me just writ this This is u this is u, this u now, this is u, u, u, u and u so this - if we distribute the cosine of u, this becomes cosine of u - let me put the y out front. so it's y cosine of u plus z cosine of u is equal to xyz sine of u. that's this first equation this second equation this second equation it looks like - well if we multiply both sides of this equation by yz we're gonna have xyz sine of u on the right hand side which is the exact same thing we have here. so let's multiply both sides of this equation by xyz, both sides. the right hand of the equation - no, not by xyz, just by yz So we are gonna multiply both sides of this equation by yz. so that these leave the denominator we are also gonna multiply the right side by yz the left hand side becomes xyz sine of u and let me write it over here just write it like this. So this is x yz sine of u and then yz times two cosine of u is going to be two z cosine of u so two z the cosine of u, I just swap the sides and then yz times two sine of u is going to be two y sine of u so two y - two y sine of u. So this is the second equation. now they don't look bad different when they were written like this it look very different. I'll thinks about this one. this side has xyz sine of u I'll do it in magenta this equation xyz sine of u.I'll write it on this side xyz sine of u is equal to - is equal to let's see we have a two z Let's distribute this cosine of u. we have a two z cosine of u so plus two z cosine - two z cosine of u plus y cosine of u so we have a plus y cosine of u. that's that times that plus y sine of u so let me scroll to the left a little bit. y sine of u so I have rewritten these three equations. The problem looks a lot less daunting right now. Let's try to figure out the number of Us between zero and three pi that will satisfy - that will give us a solution here. so let's see all of these three equations are equal to this expression right over here. so the left-hand sides - the left-hand sides of these equations all have to equal each other because they're all equaled the exact same value so let's do that. Let's see what we can do in the way of canceling in the way of canceling things out. Let's see if we use this is a plus right here. I don't know I wrote an equal here so let's see if we write this has got to be equal to that. So let me use these two first so this thing so we have two y sine of u plus two z cosine of u is equal to this which is that, which has to be equal to this, is going to be equal to y sine of u plus y cosine of u plus two z cosine - plus two z cosine of u we have two z cosine of u on both sides, so that gets rid of the z terms and then we have a two y sine of u and we have a y sine of u. So if we subtract y sine of u from both sides we end up with a y sine of u y sine of u to subtracting this from both sides,two y minus y sin of u it's just gonna be y sine of u is equal to y cosine of u. so in order for this, in order to have a solution here, in order for this to be a [inaudible] statement. remember y cannot be equal to zero in order for this to end up having a solution the coefficient on y have to equal each other. sine of u has to be equal to sine of u has to be equal to cosine of u. so that's one constraint. sine of u has to be equal to cosine of u And let's think about the unit circle and think about how many times are the sine and the cosine equal to each other when you're going between zero and three pi. So I got the unit circle right over here now clearly when we're at forty five degrees sine and cosine are equal to each other or forty five degrees is the same thing as pi over four there they're equal to each other you might be tempted to do this over here but here the cosine is negative sine is positive. so that won't work they're both negative over here but they're equal. so that's another value and then this won't work So so far we've traveled to two pi. we can go another half, we can go three pi, so we can go back to this one again so we can go back to this value again. so there are one two and then when you go all the way around again three values and then we can't go back to this one because we can only go - Let me just use another color. We can only go three pi for u. So we can only go - that's two pi and then go another time around. that is three pi so there's one two three values so there's three three possible Us. Just from this constraint, just one we use this blue equation and this magenta equation. Now let's just make sure - Let's just make sure that there aren't any further that there aren't any further constraints over here. so let's see if we can - let's use two of the other equations and the one that I would want to use that seems like there might be some cancellation. Well we can just use - We could use either this guy and this guy y cosine of u plus two z Yeah. Why not? So let's use - So as long as we're using all three equation in our constraints. we will have kind of properly constrained all of the possible solutions. and so let's think about this. This equals that, which is equal to that, which is equal to that. so we can write y cosine of u plus z cosine of u is equal to this whole thing over here, is equal to y sine of u plus y cosine of u plus two z cosine of u and then we have a y cosine of u on both sides. Those would cancel out. we can subtract a z cosine of u from both sides. a z cosine of u from both sides and so we would get zero is equal to y sine of u plus z cosine - plus z cosine of u and this seems like a pretty benign statement. y sine of u plus z cosine of u is equal to zero let me make sure this. y sine of u plus z cosine of u is equal to zero that means that this is means that two times this is also going to be zero. so this is equal to two y sine of u plus two z cosine of u is equal to zero. The only reason is I multiply it by two because by now this looks identical to this so - when you're used this equation and this equation, I got a constraint that this expression right over here this expression right over here essentially needs to be equal to zero that this expression over here essentially needs to be equal to zero which is ok which is ok because u could make the sine equal zero or x can be equal to zero. remember, they didn't put any constrains on x x can be equal to zero so this is really constraining us. x can clearly be equal to zero which could make this thing equal to zero. so it's not limiting our constraints. so now we've used all of the information that's in the problem. we've used all three of these surfaces essentially to find out if there's an inter - what the constraints we have for intersection of the three surfaces and the only real constraint is that the sine of u has to be equal to the cosine of u and there're three possible Us between zero and three pi that satisfied that.