Maybe Euler's formula may be useful

[hedlund]

Prove that if $$y = \sin^n{x}$$ then the only value for which $$y' = \sin{nx}$$ is for n=2. I'm think maybe Euler's formula may be useful ... $$\sin{x} = \frac{e^{ix}-e^{-ix}}{2i}$$ but I really got no good idea on how to solve it.

 

[Muzza]

If y = sin^(n)(x), then y' = nsin(x)^(n - 1)cos(x). Suppose n is odd and that y' = sin(nx). Then sin(nx) = nsin(x)^(n - 1)cos(x) even for x = pi/2. The equation reduces into

sin(n * pi/2) = nsin(pi/2)^(n - 1)cos(pi/2)
<=>
sin(n * pi/2) = 0.

But n = 2k + 1 for some integer k, so sin(n * pi/2) = sin( (2k + 1)pi/2 ) = sin(kpi + pi/2) = cos(k * pi). But that is never equal to zero. Contradiction.

I don't know how to handle the case when n is even and > 2.

 

[M98Ranger]

Euler's formula, which states that e^(ix) = cos(x) + isin(x), can indeed be useful in solving this problem. Let's start by substituting Euler's formula into the equation for y:

y = (e^(ix) - e^(-ix))^(n) / (2i)^n

Next, we can expand this using the binomial theorem:

y = (e^(inx) + (-1)^n e^(-inx) - 2C1 e^(i(n-1)x) + 2C2 e^(i(n-2)x) - ... - 2C(n-1) e^(ix)) / (2i)^n

Now, we can take the derivative of both sides:

y' = n(e^(inx) - (-1)^n e^(-inx) - 2C1 e^(i(n-1)x) + 2C2 e^(i(n-2)x) - ... - 2C(n-1) e^(ix)) / (2i)^n

We can see that the only way for y' to equal sin(nx) is if all of the terms involving e^(ix) cancel out. This can only happen when n = 2, as all other values of n will result in at least one term involving e^(ix) that cannot be cancelled out.

Therefore, we have proven that the only value for which y' = sin(nx) is for n = 2. This is a useful result that can be applied in numerous mathematical and scientific contexts.