IIT JEE trigonometric constraints 2010 IIT JEE Paper 1 Problem 47 Trigonometric Constraints
IIT JEE trigonometric constraints
- The number of values of theta in the interval from negative pi over two to pi over two,
- not including those two since it's got curly parenthesis around it,
- such that theta does not equal n * pi over five, so it's not a multiple of pi over five,
- for n equals zero, plus or minus one, plus or minus two, and tangent of theta is equal to
- cotangent of five theta, as well as sin of two theta is equal to cosin of four theta is ?
- We have to find the number of values of theta that satisfy all of these things.
- So just from the get go it seems we are going to have to solve for thetas and convert between sin and cosin,
- which is essentially what we are doing between tangent and cotangent.
- Those of you who know me, if you are actually taking an exam under time pressure, I don't recommend you
- reproving trig identities from first principles, but for education purposes,
- I always like to do it, so let's just draw a right triangle over here.
- If you are actually going to take the IIT JEE exam, I recommend having most of the trig identities
- at your fingertips. But let's say that's a right triangle, and let's call that theta.
- And this angle over here is going to be pi over two minus theta. If we are in radians,
- ninety degrees right here is pi over two. The whole triangle is one hundred eighty degrees which is pi,
- so this is pi over two, so these two guys are going to have to add up to the other pi over two.
- So this is pi over two minus theta. So let's think about what cosin of theta is.
- Cosin of theta. soh - cah - toa. Let's call this a, b, and c. Cosign of theta,
- it's going to be the adjacent side, let me write down. soh - cah - toa.
- Cosin is adjacent over hypotenuse. So cosin of theta is b over c, adjacent over hypotenuse.
- What is also equal to b over c? If you look from this angle over here's point of view, b is the opposite
- side. So it's opposite over hypotenuse from this angle's point of view.
- So it's also equal to the sin of pi over two, minus theta. So we've got our first
- main identity here, that the cosin of theta is equal to the sin of pi over two minus theta.
- We can go the other way around, and use the exact same logic to get that the sin of theta is equal
- to the cosin of pi over two minus theta. So that will be pretty useful when we try to solve this equation.
- We can even use it when we solve this one, because if I was to write down the cotangent of five theta.
- So the cotangent of five theta, is the exact same thing as the cosin of five theta over the
- sin of five theta. Which is one over the tangent. I can convert the cosin to sin using
- this identity over here. This is the sin of pi over two minus five theta, over the cosin of pi over
- two minus five theta. I'm just using this identity over here. This is the exact same thing
- as the tangent of pi over two minus five theta. And since we're going to have to solve this equation eventually,
- and we want to make sure we get all of the solutions to this equation,
- one thing you might want to remember is, when we take the tangent of an angle,
- and I'll just draw the unit circle here, if I have some angle, let me draw my axes.
- And let's say that this right here is theta. You probably remember that theta is essentially the
- slope of the line formed, it's opposite over adjacent, the slope of the line formed by the radius of
- the unit circle. So this theta is going to have the exact same tangent as if we add a hundred eighty
- degrees. Or if we add pi to this. So if you add pi, you get theta plus pi,
- So this whole angle over here is theta plus pi. And it's tangent is going to be the same thing.
- The slope of the radius is going to be the same. So you can add multiples of pi
- to a tangent value and you are going to get the exact same value.
- So let me put some multiples of pi over here so we can make sure that we get all of the solutions.
- So with that said, with this written, we can now solve this first equation over here.
- We can worry about some of these other constraints later. Tangent of theta is equal to
- cotangent of five theta. Well so we can write tangent of theta is equal to, and instead of writing
- cotangent of five theta we can write, is equal to the tangent of pi over two minus five theta plus
- integer multiples of pi. And so the tangent of this is equal to the tangent of that, so these things
- are equal to each other. So we get theta is equal to pi over two minus five theta plus n times pi.
- Or at minimum when you take the tangent of this it is equal to either the tangent of the same thing
- or we could add multiples of pi to it. But now we just have to solve this.
- We can add five theta to both sides of this equation. So you get six theta is equal
- to pi over two plus n pi. Plus multiples of pi. Divide both sides by six.
- I get theta is equal to pi over twelve plus n times pi over six. Or just to make the fraction easier,
- this is the same thing as pi over twelve plus two n pi over twelve.
- This is the same thing as n pi over six, I just multiplied the numerator and denominator by two.
- And this is the exact same thing as 2n plus 1 times pi over 12. So that's all of the solutions
- for this first equation over there. And let's do the same thing for this second equation
- and then we'll see where they overlap in this range, and we'll take out anything that satisfies this
- criteria right over here. So the second equation, let me write it over here, we have sin of two theta.
- Cosin of four theta, what does that equal to? Cosin of four theta using the same exact logic,
- is equal to sin of pi over two minus theta. And of course when we take a sin or cosin, or really we
- take any trig identity or any trig function of any angle, you're going to get the same
- value if you add multiples of two pi. So let me add multiples of two pi here, just so we make sure
- that we can get all the possible solutions. So this is what cosin of four theta is equal to.
- Let me be careful. Cosin of four theta is equal to sin of pi over two minus FOUR theta plus two pi n.
- You add multiples of two pi you're going to go back to the same angle. So you go back up to this equation
- up here, you get sin of two theta, is equal to cosin of four theta, which is equal to this over here,
- sin of pi over two minus four theta plus two pi n. And so these evaluate to the same thing,
- let's set them equal to each other. Two theta is equal to pi over two minus four theta plus two pi n.
- Let's add four theta to both sides of this equation, we get six theta is equal to pi over two, this is cancelled out,
- plus two pi n. Let's divide both sides by six, we get theta is equal to pi over twelve plus pi n over
- three. Now we can put it over a common denominator just to simplify things.
- We have over twelve, we have this pi. Pi n over three is the same thing as plus 4 n pi over twelve.
- Or we can write this as being, this is equal to four n plus one times pi over twelve.
- So now we just need to see where there's overlap between this solution and that solution.
- Remember we just have to count the number of solutions, we don't have to find solutions.
- And you can think about how many of these you're going to have between negative pi over two
- and pi over two, remember that was the range that we're working with, between negative pi over two
- and pi over two, not including those two. And you can figure out every one of these is also
- going to be one of these. Because no matter what you set n equal to, if you set n equal to
- twice that you're going to have equivalent things. So anything that satisfies this equation will also
- satisfy this equation right here. So really we can just count how many are in this one.
- But just to make it clear, I'm going to find all of them that satisfy this equation,
- And we're going to see how many of those satisfy this equation as well.
- But the fast way to do it is to just find or count all the ones that satisfy this one, and you've solved the problem.
- So let's just count starting at n equals zero. At n equals zero, and I'm using this n up here.
- If n equals zero we just have pi over twelve, at n equals one we have three pi over twelve.
- If n is two you get five pi over twelve. And we can't make n equal three, because
- if n was equal to three this would be seven pi over twelve which is greater than pi over two.
- That's greater than six pi over twelve so we can't have that. We can also go to negative n.
- If n is negative one, then this becomes negative pi over twelve. If n is negative two,
- this becomes negative three pi over twelve, and if n is negative three then
- this becomes negative five pi over twelve. And once again we can't go to n is equal to negative
- four because then we get negative seven pi over twelve which is out of our range.
- So this satisfies this equation up here, now which of these overlap with this over here.
- So let's set n equal to zero, we get pi over twelve. If n is equal to one we get
- five pi over twelve. We can't set n equal to two because then we get out of our range.
- We'll go above pi over two. If n is equal to negative one, we get negative three pi over twelve.
- Is n is equal to negative two then we get negative seven pi over twelve,
- which is smaller than negative pi over two so that doesn't count. So if we look over here,
- we have three solutions, and none of them satisfy this, none of them are multiples of n pi over five.
- So the answer to our problem is three. And you really could have done this second equation over here,
- and realized that anything that is a solution to this top equation is a solution to this bottom equation.
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