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Trig challenge problem: multiple constraints

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The number of values of theta in the interval from negative pi over 2 to pi over 2-- and it's not including those because we have curly parentheses around it --such that theta does not equal n pi over 5. So it's not a multiple of pi over 5, for n equals 0 plus or minus 1, plus or minus 2. And tangent of theta is equal to cotangent of 5 theta, as well as sine of 2 theta is equal to cosine of 4 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 like we're going to have to solve for thetas and convert between sine and cosine which is essentially what you're doing between tangent and cotangent. We'll do that in a little bit. And those of you who know, if you're actually taking an exam under time pressure, I don't recommend you reproving your trig identities from first principles. But for education purposes, I always like to do it. Let me just draw a right triangle right over here. If you're 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 then, this angle right over, here is going to be pi over 2 minus theta, if we're in radians. 90 degrees right here is pi over 2. The whole triangle right over here is 180 degrees, which is pi. So this is pi over 2. And so these two guys are going to have to add up to the other pi over 2. So this is pi over 2 minus theta. So let's think about what cosine of theta is. Cosine of theta, "sou-cah-toa," cosine of theta is this side. Let's call this a, b, and c. It is going to be the adjacent side. So cosine is adjacent. Let me write this down, "sou-cah-toa." Cosine is adjacent over hypotenuse. So cosine of theta is b over c, adjacent over hypotenuse, is equal to b over c. But what's also equal to b over c? If we look at this angle over here, from this angle's point of view, b is the opposite side. So it's opposite over hypotenuse for this angle's point of view. So it's also equal to the sine of pi over 2 minus theta. So we got our first main identity here, that the cosine of theta is equal to the sine of pi over 2 minus theta. And we could go the other way around, use the exact same logic to get that the sine of theta is equal to the cosine of pi over 2 minus theta. So that'll be pretty useful when we try to solve this equation right over here. And we can even use it when we solve this one. Because if I were to write down the cotangent of 5 theta-- let me write this down. So the cotangent of-- not cosine-- the cotangent of 5 theta, this is the exact same thing. The cotangent of five theta is the exact same thing as the cosine of 5 theta over the sine of 5 theta. It's just 1 over the tangent. And this is the same thing. I can convert the cosine into sine using this identity up here. This is the sine of pi over 2 minus 5 theta over the cosine of pi over 2 minus 5 theta. And for here, I'm just using this identity over here. And this is the exact same thing as the tangent of pi over 2 minus 5 theta. And since we're going to actually have to solve this equation eventually, and we want to make sure we get all of the solutions to this equation. One thing that you might want to remember is, when you take a tangent of an angle-- and I'll just draw the unit circle here-- so if I have some angle-- let me draw my axes. And let's say that this right here is theta. You probably remember that this theta is essentially the slope of the line formed. It's opposite over adjacent. It's the slope of the line formed, or by the radius in the unit circle. So this theta is going to have the exact same tangent as if we add 180 degrees or if we add pi to this. So if you add pi, you get theta plus pi. So this whole angle right over here is theta plus pi. And its tangent is going to be the same thing. It's the slope of the radius is going to be the same. So you can add multiples of pi to a tangent value, and you're going to get the exact same value. So let me put some multiples of pi over here, so that 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. And then, we can worry about some of these other constraints later. So tangent of theta is equal to cotangent of 5 theta. Well, so we can write tangent of theta is equal to-- and instead of writing cotangent of 5 theta, we can write is equal to the tangent of pi over 2 minus 5 theta plus integer multiples of pi. And so the tangent of this is equal to tangent of that. These things could be equal to each other. So we get theta is equal to pi over 2 minus 5 theta plus n pi. Or at minimum, this could be equal, when you take the tangent of this, it's equal to the tangent of either the same thing, or we could add multiples of pi to it. You could view it as either way. But now, we just have to solve this. We can add 5 pi to both sides of this equation. We get, sorry, we could add 5 theta to both sides of this equation. We're adding this to both sides. So you get 6 theta is equal to pi over 2 minus n pi, minus multiples of pi, or sorry, plus multiples of pi. There's a plus right over there. Divide both sides by 6. I get theta is equal to pi over 12 plus n times pi over 6. Or just to make the fraction adding easier, this is the same thing as pi over 12 plus 2. Let me write it this way. Plus 2 n pi over 12. This is the same thing as n pi over 6. I just multiplied the numerator and denominator by 2. And this is the exact same thing. You could imagine a 1 coefficient out here. This is the same thing as 2n plus 1 times pi over 12. So that's all of the solutions for this first equation right over there. Now, 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 sine of 2 theta. Let me write it over here first. Cosine of 4 theta. What does cosine of 4 theta is equal to? Cosine of 4 theta, using the same exact logic, is equal to sine of pi over 2 minus theta. And of course, when we take a sine or cosine or we take really any trig identity or any trig function of any angle, you're going to get the same value if you add multiples of 2 pi. So let me just add multiples of 2 pi here, just so we can make sure that we get all the possible solutions. So this is what cosine of 4 theta is equal to. So let us-- oh, let me be careful. Cosine of 4 theta is not equal to pi over 2 minus theta. That would be cosine of theta. If it's cosine of 4 theta, it's pi over 2 minus 4 theta plus multiples of 2 pi. So plus 2 pi n. Because obviously you add multiples of 2 pi, you're going to go back to the same angles. So you go back up to this equation up here. We get sine of 2 theta is equal to cosine of 4 theta. Which is the same thing as this over here, which is equal to sine of pi over 2 minus 4 theta plus 2 pi n. And so these evaluate to the same thing. Let's set them equal to each other. 2 theta is equal to pi over 2 minus 4 pi plus 2 pi n. Let's add 4 pi to both sides of this equation. We get 6 theta. 6 theta is equal to pi over 2. So this is canceled out. Plus 2 pi n. Let's divide both sides by 6. We get theta is equal to pi over 12 plus pi over 3n. 2 divided by 6 is pi n over 3. Now, we can put it over a common denominator just to simplify things. So we have over 12. We have this pi. pi n over 3 is the same thing as plus 4n pi over 12. Or we could write this as being, this is equal to 4n plus 1 times pi over 12. So now, we just need to see where there is overlap between this solution and that solution. Remember, we just have to count the number of solutions. We actually don't have to find the solutions. And actually if you pretty savvy and do things in your head, just at this point, you could think about how many of these you're going to have between negative pi over 2 and pi over 2. If you remember, that was the range that we're working with. Between negative pi over 2 and pi over 2, not including those two. And then you can figure out, and actually 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 the equivalent thing. So any of these are going to be any of these. 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 then we're going to see how many of those satisfy this equation as well. But the fast way to do it, just find the ones that satisfy this one. Or just count the ones that satisfy this one. And you've done the problem. So let's just count. So let's just start at n equals 0. If n is equal to 0-- and I'm using this n up here. I won't even write that. If n is equal to 0, we just have pi over 12. If n is 1, we get 3 pi over 12. If n is 2, we get 5 pi over 12. And we can't make n equal 3. Because if n was equal to 3, this would be 7 pi over 12 which is greater than pi over 2. That's greater than 6 pi over 12. So we can't have that. We could also go to negative n. We could also have negative n. If n is negative 1, then this becomes negative pi over 12. If n is negative 2, this becomes negative 3 pi over 12. And if n is negative 3, then this becomes a negative 5 pi over 12. And once again, we can't go to n is equal to negative 4, because then we get negative 7 pi over 12 which is out of our range. So this satisfies this equation up here. Now, which of these are overlap with this over here? So let's set n equal to 0. We get pi over 12. If n is equal to 1, we get 5 pi over 12. We can't set n equal to 2, because then we'll get out of our range, we'll go above pi over 2. If n is equal to negative 1, we get negative 3 pi over 12. If n is equal to negative 2, this is negative 8. Then, we get to negative 7 pi over 12, which is smaller than negative pi over 2. So that doesn't count. So there are three solutions. Or if we look over here, we have three solutions. We have one, two, three solutions. And none of them satisfy this. None of them are multiples of n pi over 5. So the answer to our problem is 3. And you really could have just done the second equation right over here. And realize that anything that's a solution in this top equation would be a solution to this bottom equation. And you would have just counted these solutions. And you would have been done.