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## Trigonometry

### Course: Trigonometry>Unit 4

Lesson 6: Challenging trigonometry problems

# Trig challenge problem: multiple constraints

Sal solves a very complicated algebraic trig problem that appeared as problem 47 in the 2010 IIT JEE Paper I exam. Created by Sal Khan.

## Want to join the conversation?

• Dear Sir,
About the middle of your solution, I think, we should have considered: 2(thita) = (pi/2)- 2(thita)[NOT: 4(thita)] + 2n(pi) • That would be assuming a 45deg angle: if we change the pi/2 from radians to degrees, we get 2theta = 90 - 2theta OR x = 90-x.

Sal did not randomly put in 4 where he should have put 2 - he used the given equation "sin 2theta = cos 4theta" to get a sin = sin equation. This is btwn and in the video.
• Is IIT for India only.
How hard is the IIT JEE?
Can I get admission into IIT Mumbai with 70 percent in IIT? • I agree that sin(x) = sin(x+2pi), but sin(x) = sin(pi-x) also, however those aren't taken into account in this video....???? Am I missing something? • The sin(x) DOES always = sin( pi-x ). This is simple trig identity, easy to proof. It holds for the given example of x=5pi/4, since sin(5pi/4)=-sqrt(2)/2=sin(-pi/4)=sin(pi-5pi/4). This identity indeed should be taken into account. And only by accident the solution found by Sal is correct. This is because the identity sin(x)=sin( pi-x ) does not provide any more solutions it this particular case which is considered in the film, but in other problems it can provide more solutions, and always should be taken into account. Consider for example the equation sin(x)=sin(2x). In this case Sal's method will not give all solutions to the equation, but only part of them, the rest can be found using the identity sin(x)=sin(pi-x).
• i just don't understand y we have to add npie • Is this trig trig or is this trig calculus? I'm taking trig should i worry if i don't get this? • What level math is this?? is it precalculus? Because it's way too difficult for me. • @ your saying cot 5 theta + npi = tan (pi/2 - 5 theta + n pi) which is arbitrary. I found another who agrees....
"Dear Sir,
About the middle of your solution, I think, we should have considered: 2(thita) = (pi/2)- 2(thita)[NOT: 4(thita)] + 2n(pi)"
2 Votes • Flag 8 months ago by a_ashkenazi
(1 vote) • Because cot = cos/sin and tan = sin/cos, if you do cot of one angle (of a right triangle) then it is the same as the tan of the other angle. Another way of writing this is:
cot(x) = tan (90-x) for degrees or cot(x) = tan (pi/2 - x) for radians. Then substitute 5theta for x.
We add the n*pi because at every cycle +pi tan (and cot) would be the same. If the range is limited to -pi/2 to pi/2, why would we add the n*pi? If we go back to graphing trig functions and graph sin(5theta) for example, we'll see 5 cycles of sin btwn 0 and 2pi instead of one. So tan(theta) = cot(5theta) and sin(2theta) = cos(4theta) means we'll have to consider the wider range until we get our final answers. If you were to graph them each (sin, cos, tan, cot, continuously both directions) then you'd see the 3 points in the given range at the end.
• So, I know that in chemistry, the units can usually be treated like variables. Is this the same for trig functions? At , it looks almost like Sal divided both sides by the tangent. Is that what he did, or was he just concluding that (theta) and (pi over 2 + 5 theta + n theta) must be equal to each other because their tangents were equal to each other? • why is that only 3 values of theta? instead of 9 since 6 for the tan theta = cot 5theta and 3 for sin 2theta = cos 4theta. I think what we really want to determine is that the number of common values of 2 equations subject to restriction.
(1 vote) • 𝜃 ∈ {−5𝜋∕12, −3𝜋∕12, −𝜋∕12, 𝜋∕12, 3𝜋∕12, 5𝜋∕12}
These six values are the only values in the interval (−𝜋∕2, 𝜋∕2)
that also solve the equation tan 𝜃 = cot 5𝜃

– – –

Similarly, we have three values,
𝜃 ∈ {−3𝜋∕12, 𝜋∕12, 5𝜋∕12},
that are in the interval (−𝜋∕2, 𝜋∕2)
and solve the equation sin 2𝜃 = cos 4𝜃

– – –

Now, the second constraint is that both equations must be true.
𝜃 ∈ {−5𝜋∕12, −𝜋∕12, 3𝜋∕12} don't meet this constraint because they only solve one of the equations.

So, that leaves us with the three values:
𝜃 ∈ {−3𝜋∕12, 𝜋∕12, 5𝜋∕12}

– – –

There is also a third constraint that says that none of these values can be a multiple of 𝜋∕5, but since none of them are we're still left with three values. 