Trigonometric system example 2010 IIT JEE Paper 1 Problem 55 Trigonometric System
Trigonometric system example
- 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.
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