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## Trigonometry

### Course: Trigonometry>Unit 4

Lesson 6: Challenging trigonometry problems

# Trig challenge problem: system of equations

Sal solves a very complicated algebraic trig problem that appeared as problem 55 in the 2010 IIT JEE Paper I exam. Created by Sal Khan.

## Want to join the conversation?

• In the actual JEE question paper, which is of three hours, i figured i'd be getting hardly two minutes for each question. For a question of such sort, which is monstrous by the very look of it, i'm not sure if i'll work it out for 'eleven' minutes, or i'll just skip it and save it for the last. Is there a 'key' idea to this which makes the problem clearer, or can it also be done by trial and error from the options ?
• In IIT, for this problem, you are thinking mentally, and not speaking or explaining to anyone. Your Mental Processes are Wayyy faster, so even if u might use Sal's Method, it wouldn't take you more than 3 mins to solve. However i analysed the q, and think that you can slightly do it faster by equating all three equations with each other one at a time, which quickly leads you to Tan 3(theta) = 1 , and the other two satisfy each other, i tried out the problem on my own and solved it under three minutes. So dw, it shouldnt take you Eleven Minutes in JEE. Hope this helped :)
• What do the little zeros under some numbers he calls nots mean?
• x_0 or x naught refers to a specific value of the variable x. It is used so that they don't need to introduce extra variables to signify an exact value.
• How were the trigonometric functions discovered?
What's the story behind this?
• The study of trigonometric functions was started by the Greeks and the Arabs. They started studying the plane and spherical angles observing the stars. They were aware of most of the theorems we now use as you can see here http://www-history.mcs.st-and.ac.uk/HistTopics/Trigonometric_functions.html
They didn't talk about sin or cos but they measured those distances with cords and then compared them with the circumference.
• How do the sine, cosine and tangent trigonometric functions work?
I mean we know that a function f(x) is the output and x is the input.
I'm asking what goes on with trigonometric functions each time we input an angle in radians or in degrees.

For instance sin90°=??unknown process??=1
I want to know for instance in this example how did sin90° become 1.
What was the process behind this?
(1 vote)
• Method I: Draw an unit circle, draw the angle, see coordinates of the point where the angle's side intersects the circle.

Method II: With calculus, it can be shown that for x in radians, the series expansion for sin x = x - x^3/3! + x^5/5! - x^7/7! + x^9/9 - ... The sum is infinite, but a finite amount of terms can still get you a good approximation. (The process is rather complicated: it requires knowing limits and derivatives.)

Method III: With complex numbers, it can be shown that sin x = (e^(ix) - e^(-ix)) / (2i). This one actually builds on method II: it requires knowing the series expansions for e^x, sin x and cos x.
• At , assuming the LHS of the purple and blue equations to be equal, can we say that y cosu = y sin u and therefore u = pi/4 ?
• Yes, in the video Sal said u = pi/4 is one of the 3 possible solutions. The other 2 possible solutions are 5pi/4 and 9pi/4. Since u could go from 0 to 3 pi (remember, at the beginning, Sal used u to replace 3 theta, to simplify the problem).
• So how can I solve the following: (Cos[x]+Sin[x])^(Sin[2x]+1) = 2 For all values of x in [0, 2Pi]
• If sin(2x) + 1 = 0, this equation will have no solution. Therefore, sin(2x) != (not equal to) -1, which gives x != 135 degrees or 3 * pi / 4 radians. But if sin(2x) + 1 = 2, this equation may have a solution. Let's try. sin(2x) + 1 = 2 => sin(2x) = 1 => 2x = 90 degrees (in the specified range) => x = 45 degrees. From this we can say that cos(x) + sin(x) = sqrt(2) put the value of x in this equation, and the equation will be satisfied. You can judge all this if you know the range of these trigonometric functions. 45 degrees is one value which satisfies this equation. However, there may be other solutions.
• At around , why can you only go 3 pi for u? Is it because the degree is 3 theta? If so, then why? I still don't get it...
• Multiplying all 3 sides by 3, we get 0<3 theta<3pi. So here 3 theta is denoted as 'u'. so 0<u<3pi.
• This isn't related to the video, but I head of "the trigonometry challenge", and I am wondering where it is. Now that I have done all of the trig on Khan, I thought that I would find it, but I haven't. Does anyone know where it is?
• Well, these videos are Trigonometry challenge problems extracted from IIT JEE 2010 Paper 1. I haven't heard about any Trigonometry Challenge so far.
(1 vote)
• At , we get that xyz*sin(u)=0. Sal explains how x can be equal to 0, but can't sin(u) also equal 0? And if u is in the interval from 0 ti 3pi, sin(u)=0 when u=pi,2pi. Shouldn't this add two more possibilities for u?
(1 vote)
• Actually, Sal says that either sin(u) or x can equal to zero. However, at , Sal begins to set the first constraint, that `sin(u) = cos(u)`. This means that you cant consider π and 2π to be values for u, as `sin(π)` is not equal to `cos(π)`, nor is `sin(2π)` equivalent to `cos(2π)`. Hope this helps!