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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?
• 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
R*sin(x+p)

You probably would see directly that R = a^2 + b^2 and p = arctan(a/b)
• :
cos^2 theta = 1+cos2 theta/2
how is this equal?
• 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
• 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.
• 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?
• 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.
• can i have the link to where sal proved cos^2 theta=(1-cos 2 theta)/2
• can we solve the given question without using derivatives.
• 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?
• 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.
• 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 :)
• Can somebody explain what taking the derivative helps Sal achieve? I'm not on calculus yet.
• 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