IIT JEE Trigonometric Constraints 2010 IIT JEE Paper 1 Problem 47 Trigonometric Constraints
IIT JEE Trigonometric Constraints
⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.
- 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.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
Have something that's not a question about this content?
This discussion area is not meant for answering homework questions.
Share a tip
When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
Have something that's not a tip or feedback about this content?
This discussion area is not meant for answering homework questions.
Discuss the site
For general discussions about Khan Academy, visit our Reddit discussion page.
Flag inappropriate posts
Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.
- disrespectful or offensive
- an advertisement
- low quality
- not about the video topic
- soliciting votes or seeking badges
- a homework question
- a duplicate answer
- repeatedly making the same post
- a tip or feedback in Questions
- a question in Tips & Feedback
- an answer that should be its own question
about the site