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Solving sin(x)^sin(x)=2

blackpenredpen

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We have two exponential equations with trigonometric functions (sin(x))^(sin(x))=2 and (sin(x))^(cos(x))=2. The tetration equation (sin(x))^sin(x)=2 requires us to use the Lambert W function, complex exponential, complex logarithm, and the quadratic formula. However, for (cos(x))^cos(x)=2, we can argue it has real solutions by intermediate value theorem but I do not know a way to solve it algebraically. I know we can use Newton's method to get an approximation for the solution but I wonder if we can somehow also use the Lambert W function to find a closed-form of the solution.

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0:00 I have a math conundrum
0:12 solving (sin(x))^sin(x)=2
7:40 why (sin(x))^cos(x)=2 has real solutions
10:16 can WolframAlpha solve (sin(x))^cos(x)=2?

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20 Comments

@blackpenredpen Creator 4 years ago Pinned

What do you think about my "e-hoodie"?

@johndoe9659 4 years ago (edited)

Numerical approximations are, to a mathematician, what frozen ready meals are to a decent chef.

@HungNguyen-rj3ek 4 years ago

It's actually okay to have an irrational number that cannot be explained in known functions and constants.
To solve this equation, just make up a new function like Larmbert W for it.

@gabequinn9796 4 years ago

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@gld67k 4 years ago (edited)

My teacher always said "When you got a difficulty with algebra, use calculus to prove the sum"

"When you got a difficulty with calculus, use algebra to prove the sum"

"When both don't work, use trigonometry and get over it"

@wesleydeng71 4 years ago (edited)

Well, in the spirit of W Lambert function, let's define a Q function such that Q(x) is the solution of sin(t)^con(t) = x. Then obviously, the solution of the equation is Q(2).

@yoav613 4 years ago

If you write in wolfram x^-sqrt(1-x^2)=2 you get the numeric sol x=0.4584 so x=arcsin(0.4584)=2.6653+2npi

@hotlatte1222 4 years ago

It is easy. I can solve it. I have a proof of my theorem, but there is not enough space in this margin.

@kelpyg3718 11 months ago

7:18 the inside of the logarithm can be simplified because it has the exact form of cosh^-1(x)

so it will be:

x = pi/2 - icosh^-1(e^w(ln2))

@goliathcleric 4 years ago

I haven't solved the second one (it's 5am and I just woke up) but my instinct is telling me the first step is going to be to convert both sin(x) and cos(x) to their complex equivalent using e. I think it'll be similar to cubic equations, where you have to journey through the complex world to find their real solution.

@darkxify 9 months ago

If anyone was wondering, when i put the sinx^cosx into desmos regression, I get x=2.66536 (radians), which seems to be approximately right.

@wiseSYW 4 years ago

you'll need to define a new function like lambert W, I guess

using google, putting in (sin x) ^ (cos x) and moving the mouse at the graph, you'll get x = 2.666... when y is about 2.0003...

@vanderavongola 4 years ago

hi @blackpenredpen! This is such a great video! I solved this equation almost the same way as you did but I had a different approach starting from 4:40. I expressed e^ix as cos x + i sin x and grouped real and imaginary terms. What I got was

x = sin^(-1) {exp[W(ln2) \pm \sqrt(exp[2*W(ln 2)]-1)}

which is a real solution if exp[2*W(ln 2)]-1. Is this consistent with your solution? Hope you see this!

@jeremymwilliams 4 years ago

Numerical Method seems to be the only way.

@arimermelstein9167 4 years ago

It’s a pain the butt, but you could use Newton’s method or bisection to find an approximation to the real solution. I’m not sure how to do it analytically either.

@chessematics 4 years ago (edited)

Little fact: e^(W(x)) = x/W(x). So one more step of simplification.

@MyRandomVideos4 2 months ago (edited)

In fact, (sinx)^(cosx) = 2 has the real solution: x = 2.543 (approx.). I used the Newton-Raphson method for that.

@ubern3rd 4 years ago

I was watching this at work and decided to try something on the second one. Instead of (sin(x))^cos(x) = 2, why not do (sin(x))^cos(x) = e? Since 2 is just a real number, we can use another real number to understand what it's doing, also I'm not very familiarized with the Labert W function with the exception that it gives you the fish back. I get stuck, you'll see where, but I wanted to lay out my thought process to see if anyone had any thoughts on it.

Start: (sin(x))^cos(x) = e
Take the cos(x) root of both sides: sin(x) = e^(1/(cos(x)))
Divide by cos(x) on both sides: sin(x)/cos(x) = (1/cos(x))*(e^(1/(cos(x))))
We have a tangent function on the left with a n*e^n expression on the right! (I knew that based on the wolfram graph that this looked kinda like a tan function).
Simplify: tan(x) = (1/cos(x))*(e^(1/(cos(x))))
Lambert W Function: W(tan(x)) = 1/cos(x)

I don't know how to simplify W(tan(x)) further, so I guess that's where I stop. If we can find this out, then theoretically, we can do it for (sin(x))^cos(x) = 2, just convert 2 to e^n and go from there? Just a thought and any thoughts on this would be cool. I'd love to see bprp solve for this :)