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Trig challenge problem: maximum value
Sal solves a very complicated algebraic trig problem that appeared as problem 48 in the 2010 IIT JEE Paper I exam. Created by Sal Khan.
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- What is the R-Formula that skywalker94 refers to in the comments?(7 votes)
- When we have an expression in the form a*cos(x) + b*sin(x),
a and b can be written as,
a = R*sin(p), for some constants R and p
b = R*cos(p)
This makes the expression, R(sin(p)*cos(x) + sin(x)*cos(p))
Which can as well be written as
You probably would see directly that R = a^2 + b^2 and p = arctan(a/b)(9 votes)
cos^2 theta = 1+cos2 theta/2
how is this equal?(6 votes)
- cos(2x) = cos^2 (x) - sin^2 (x)
= cos^2 (x) - sin^2 (x) - cos^2 (x) + cos^2 (x)
= 2cos^2 (x) - 1 [as sin^2 (x) + cos^2 (x) = 1]
so cos^2 (x) = (1 + cos(2x) )/2.
Hope it helps(6 votes)
- Here's an easier way. Once the denominator is in the form
3+Acosx+BSinx, you just realize that ACosx+BSinx is a
sine wave with amplitude (A^2+B^2)^(1/2) which is plus or minus 5/2. Then the minimum value of the denominator is 3-5/2 or 1/2, giving the expression a maximum value of 2.(8 votes)
- Where does this identity come from? Seems very useful, I don't quite understand it.(1 vote)
- In this video, Sal used a Calculus operation. I don't know what a derivative is, so how am I supposed to understand that part of the video? Is there a simple way of explaining a derivative that I would understand?(5 votes)
- The derivative of a function is usually another function that describes the rate of change of the original function. It is basically slope generalized to any curve. The derivative of a line would be its slope but how do we find the rate of change of curves like parabolas? This is where differential calculus comes in and where we use derivatives. For a more detailed explanation, I would look into differential calculus videos on KA.(4 votes)
- can i have the link to where sal proved cos^2 theta=(1-cos 2 theta)/2(4 votes)
- here'sthe link you asked for:
- I see that one of the ways to solve this problem is through the Harmonic Addition Theorem, but how exactly does that theorem work? More specifically, what are the applications of the dirac delta function, which is featured prominently in the theorem?(3 votes)
- We changed two times theta to x, and never turned it back into the original form. Why does this not change the value of the answer? Thanks in advance.(2 votes)
- Since we are not calculating the actual value of 2(theta) or x, this does not matter. The value of sin 2(theta) = sin x ,since we assumed that 2(theta)=x . Unless u are trying to solve for theta, this conversion doesnt actually change anything. The reason Sal changed from 2(theta) to x is so that it is easier to understand the part of finding the derivative (i think :P).......anyway....hope that helped :)(2 votes)
- can we solve the given question without using derivatives.(2 votes)
- Can somebody explain what taking the derivative helps Sal achieve? I'm not on calculus yet.(2 votes)
- we take derivatives to find the points where the slope is zero. this gives what is called the local maximum and local minimum (the highest and lowest values for f(x)).
Take the graph for f(x) = sin(x).
f'(x)=cos(x), cos(x) = 0 at x = pi/2 and 3pi/2.
this is where sin(x) is greatest and least.
The points where slope is zero is when sin(x) is either at its highest or lowest.
Sal wanted to find the minimum of the denominator to find the maximum of numerator. So he found the minimum using derivatives.(1 vote)
- if cosx-sinx=1 ,then how can i prove cosx+sinx=2^1/2(1 vote)
- squaring both sides of cosx-sinx=1 we get 2sinxcosx=0 . again (cosx+sinx)^2=1+2sinxcosx=1+0
hence cosx+sinx=1(2 votes)
The maximum value of the expression 1 over sine squared of theta plus 3 sine theta cosine theta plus 5 cosine squared of theta is? So let's rewrite this. So this is 1 over, so I have a sine squared of theta. And then, at least in my brain, whenever I see a sine squared of theta, I always look for a cosine squared of theta. Because I know that when I take the sum of them it equals 1. And I don't have just one cosine squared theta here, I have five cosine squared theta. So let me just take one of them. So I have a plus cosine squared theta. And since I took one of them, I only have four of these left. So plus 4 cosine squared theta. And then I have this stuff. Plus 3 sine of theta cosine of theta. So what this first step allowed me to do is just turn these two characters right over here, sine squared theta plus cosine squared theta, that is equal to 1. So we've simplified it to 1 over 1 plus-- now let's think about how we can write cosine squared of theta. I'll write our identities here. Cosine squared of theta. And we've prove these in the trigonometry playlist. This is equal to 1 plus cosine of 2 theta all of that over 2. And my goal here is, I really just want to get everything-- well, I really just want to simplify it. And maybe we'll do calculus. We'll use a little calculus to find the minimum value of the denominator, which would give us the maximum value of the numerator. So let's see, cosine squared of theta is equal to this. So 4 times this is just going to be-- so 4 times this. The 4 divided by 2 is 2. So it's going to be 2 times this numerator. So it's going to be 2 plus 2 cosine of 2 theta. That's this term right here. And then this term right over here, we could use the trig identity. That sine of two theta is equal to 2 sine of theta cosine of theta. Or you divide both sides by 2, you get 1/2 sine of 2 theta is equal to sine of theta cosine of theta. So this right over here, this part right over here is going to be 1/2 sine of 2 theta. But we're multiplying it by 3. So it's going to be plus 3/2 sine of 2 theta. And let's see, this part over here clearly simplifies. This is 3. So let me just rewrite it. This is 1 over 3 plus 2 cosine of 2 theta plus 3/2 sine of 2 theta. And we're really just looking for the minimum. We're looking for the minimum value of the denominator, which would give us the same as the maximum value of the numerator. It would just be 1 over this minimum value. So let's see how low we can get, assuming we're above 0. Let's see how low we can get for this denominator right here. We're just going to look for its minimum value. So one thing we can do, just to simplify things, the minimum value of this is going to be the same. The min of this thing-- I don't want to write it there because it kind of confuses the problem. The minimum value of 3 plus 2 cosine 2 theta plus 3/2 sine of 2 theta is going to be the same thing as the minimum value of 3 plus-- I'm just going to do the substitution that 2 theta is equal to x. That just simplifies things a little bit. You don't have to do that. But just to, so 3 plus 2 cosine of x plus 3/2 sine of x. So this is a pretty simple expression. Let's see how we can figure out its minimum value. And my temptation is to take the derivative, find out where the derivative is equal to 0, and then that will either be a maximum or a minimum point. So let's take the derivative. The derivative of this expression right over here with respect to x. Well, derivative of 3 with respect to x is 0. Derivative of cosine of x is negative 2 sine of x. Derivative of 3/2 sine of x is going to be plus 3/2 cosine of x. And then, that is going to be equal to 0. We want to find where the slope is 0 because it's either going to be a maximum or minimum point. And let's see, we can add, we could add 2 sine of x to both sides. So we get 3/2 cosine of x is equal to 2 sine of x. And then, we can divide both sides of this equation by, let's divide it by 2 first. I don't want to skip too many steps. So 3/4 cosine of x is equal to sine of x. Then we could divide both sides by cosine of x. So we get 3 over 4 is equal to sine of x over cosine of x, which is the same thing as the tangent of x. So an x value that gives us-- or the tangent of an x value that gives us 3/4 is going to give us either a maximum or a minimum point. So let's think about this. Let's think about this a little bit. Let me draw my unit circle. So let's think about two x values that will give us the tangent of 3/4. So let me draw my unit circle. That's a unit circle. Let me draw a unit. This is always the hardest part. So let me draw this. All right, there is my unit circle. So how can I get a triangle, or, well, yeah, let's think about that way. How can I get a triangle where an angle is the tangent of 3/4. Remember, tangent is opposite over adjacent. Right? Tangent is opposite over adjacent. So if this is my triangle right over here, if this is x, opposite over adjacent is equal to 3/4. So opposite could be 3 and adjacent could be 4. And we hopefully immediately recognize this. This is a 3, 4, 5 triangle because it's a right triangle. 3 squared plus 4 squared is 25 which is 5 squared. So this is a 3, 4, 5 triangle. Now, there's two tangent values. So x could be like this. So this obviously isn't a unit hypotenuse over here. But we could divide everything by 5, and it would be. So we could have this situation. We could have this situation over here. Where this is x. If this is the unit circle, the hypotenuse is 1. This is 3/5 and this is 4/5. This would, tangent of x here, tangent of x would give us 3/4. But is this going to give us a maximum value or minimum value? Well over here, both cosine of x and sine of x are going to be positive. So both of these are going to be positive values. So it's going to, probably, maximize this expression over here. Now the other x that gives us the same tangent, and remember, the tangent is really just the slope of the radius in the unit circle, would be this angle. This has the same tangent value. So this x over here. In this case, the tangent is still going to be 3/4. But over here, the sine and cosine are negative. So over here, the x-coordinate, or the cosine, is going to be negative 4/5. And the sine value, or the y value, is going to be negative 3/5. And this will give us a minimum point, because here both the sine and the cosine are negative. So let's use this x right over here. And notice, we don't even have to figure out what the x is. Because we know that if the tangent over here is 4/5, either both the sine is going to be 3/5, and the cosine is going to be 4/5, which will give us a maximum point. Or the tangent could be 3/4. And then the sine will be negative 3/5, and the cosine will be negative 4/5. So let's use these over here. So the minimum is going to be equal to 3 plus 2 times cosine of x. We're using this one over here. So 2 times cosine of x. Cosine of x here is negative 4/5. And then plus 3/2 times the sine of x. Sine of x here is negative 3/5. And what is this going to be equal to? This is going to be equal to 3 plus, this is negative 8/5, 3, maybe I should write 3 minus 8/5 minus 9/10. And so this is going to be equal to-- we could put everything over 10 --30 over 10 minus 16 over 10, that's 8/5. Minus 9/10. And this gives us what? This gives us 5/10. 5/10 or 1/2. So the minimum value for our denominator-- everything we've been dealing with so far has been our denominator-- the minimum value of all of this business over here, the minimum value is 1/2. So the maximum value, that this whole expression takes, is when the minimum value is 1/2. So we get 1 over 1/2 half which is equal to 2. And we're done.