solving tan(1/x)=1/tan(x)

A viewer wanted me to try the trig equation involving tangent: tan(1/x)=1/tan(x). This seems to be a fake trig identity but it turns out to be a very interesting trig equation. We will use some trig identities to simplify this equation and also the quadratic formula to actually solve this equation.

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[mikefochtman]

"I know because it's on my board, it must be true." lol Not sure if my prof would have accepted that particular proof, but loved it. :)

[jozsefgurzo]

Very nice problem and a cool video! I would just like to note that since we assume that x is a real number, n cannot be 0, or -1, because we have (2n+1)^2 π^2 - 16 under the square root, which is negative in these cases.

[intraced]

I love how any equation where you apply the same function to both the input and main function/operation always has such a good answer and explanation

[TheDannyAwesome]

Great equation. My first instinct was to write both tangents as sin/cos, rearrange, and use compound angle formula for cosine. Your method looks neater, but you have to be careful with the tan(a)=tan(b), whereas I only needed to deal with cos(a)=0.

[l_szabi]

First, multiply by tanx
tanx * tan(1/x) = 1
then:
(sinx * sin(1/x))/(cosx * cos(1/x)) = 1
multiply by the denominator:
sinx * sin(1/x) = cosx * cos(1/x)
subtract the rhs:
sinx * sin(1/x) - cosx * cos(1/x) = 0
use the sum identity:
-cos(x + 1/x) = 0
therefore:
x + 1/x = pi/2 + -2*k*pi- k*pi
and the same from here.

_edit: corrected 2*k*pi_

[Dreamprism]

1:11 I'm always telling the students I tutor what "co" means because somehow teachers and textbooks don't often make it clear enough. And students often forget the co-function identities to boot!

[bodyrockdance]

I got to the quadratic a little differently, and I think it's pretty cool. If you multiply through by tan(x), you end up with the equation tan(1/x)*tan(x) = 1. What THAT means is that tan(x + 1/x) must be undefined, due to the denominator of the tan compound angle identity. Therefore x+ 1/x = pi/2 +n*pi, as this is the set of values for which tan is undefined.

[Inspirator_AG]

Freshmen's dream, but level 15 septillion.

[Ninja]

The complementary identities for trigo functions are underrated imo. Most of my classmates dont even know the one with sine and cosine

[SyberMath]

I didn't even know you made a video of this. Good work!!! 🤩🤩🤩

[johnchessant]

sin(1/x) = 1/sin(x) : If we're looking for real solutions only, then we can use the fact that |sin(x)| <= 1, which means |1/sin(x)| >= 1, so the only way sin(1/x) could equal 1/sin(x) is if both were equal to +/- 1. But the solutions to 1/sin(x) = +/- 1 are x = n*π and the solutions to sin(1/x) = +/- 1 are x = 1/(n*π). They have no solutions in common, so sin(1/x) = 1/sin(x) has no solution.

If we're looking for complex solutions, wolfram alpha gives 4 solutions, x = +/- 0.719 +/- 0.695 i. I noticed these solutions have |x| = 1 so we can set x = exp(i*t) and solve sin(exp(-i*t)) = 1/sin(exp(i*t)). This can be rewritten as cosh(2 sin t) - cos(2 cos t) = 2, which means I don't think there's any hope of expressing the solutions in closed form.

[threepointone]

Idea: prove Σ(n = 1, ∞) of n^(2k) is always 0 for k is an integer. Tip: use ramanujan summation f(0)/2 + Σ f(n) = i ∫ (f(it)-f(-it))/(e^(2πt)-1) dt

[enartogo]

4:30 all equal to pi lmao, I love the confidence with which he says it although there’s a 0 there

[zackcarl]

Thank you for explaining the meaning of the cotx now i don't have to mug up the identity i understand the meaning why we take cot x = tan(90-x), i have studied that a complementary angle is when two angles add up to 90*, but my teacher never explained me the meaning of cot x is and how to connect the two concepts

[samvergolias]

4:32 “We all make mistakes in the heat of passion, Jimbo”

[almightyhydra]

1:41 we are all enlightened adults now so we will say tau/4 :p

4:33 particularly if pi = 0

[georget]

I used the euler's formula to calculate tanx as sinx/cosx. Then I applied simple algebra and I ended up with x=-[π+-sqrt(π^2+16)]/4

[EDoyl]

some advanced in-your-head math here. it's not easy to have above and below factors in your head all at once, but it's good to practice. nice video.

[Wmann]

I just love it when he says “Now this is soo cool, because…”, because it’s so cool.

[brridk]

i could have definitely used this in my pre-calc class what!! this was explained so well, thank you