# What real $\lambda$ satisfy $\frac{\sin\lambda\alpha}{\sin\alpha}-\frac{\cos\lambda\alpha}{\cos\alpha}=\lambda-1$?

$$\frac{\sin \lambda \alpha } {\sin \alpha} - \frac{\cos \lambda \alpha}{\cos \alpha} = \lambda - 1$$ where $\alpha$ cannot be integral multiple of $\frac{\pi}{2}$.

Playing with the LHS I ended up with following equation $$\frac{2\sin \{\alpha(\lambda - 1 )\}}{\sin 2\alpha } = \lambda -1$$ after that i don't how to solve this trig equation.

• This question is ambiguous. See the discussion with Dylan under my answer. Do you mean, "For what values of $\lambda$ is this an identity for $\alpha \ne k\pi/2?$" or do you mean, "For what values of $\lambda, \alpha$ does the following equation hold?" – saulspatz Feb 23 '18 at 20:12

By inspection, $\lambda = 1$ is a solution. When $\lambda \ne 1,$ we have $2\sin(\lambda-1)x = (\lambda-1)\sin(2x)$. Taking the Taylor series of both sides, $$2(\lambda-1)x - 2\frac{(\lambda-1)^3x^3}{3!} + O(n^5)= (\lambda-1)(2x)- (\lambda-1)\frac{2^3x^3}{3!}+O(n^5)$$ These must be equal term-by-term. The first terms are identical. Comparing the second terms, we see that $(\lambda-1)^2 = 4 \implies \lambda-1 = \pm 2\implies \lambda \in \{-1,3\}.$ We easily check that these give solutions, so the only solutions are $\boxed{ \lambda \in \{-1,1,3\}}$

• What if $x = k\pi/2$? – Dylan Feb 23 '18 at 19:02
• there is $\cos \alpha$ in the denominator in the equation given in the question so this would make the term undefined at $x= \frac{k \pi } {2}$ – Ranjeet Bahadur Feb 23 '18 at 19:18
• That's a good point, but there are other cases. Refer to my answer. – Dylan Feb 23 '18 at 19:39
• @Dylan This works for all values of $x.$ The excluded values of $x$ come from the form of the original question, to avoid dividing by $0$. – saulspatz Feb 23 '18 at 19:46
• You do get additional solutions if $\frac{d}{dx} \left( \frac{\sin 2x}{2x} \right) = 0$ – Dylan Feb 23 '18 at 19:49

It helps to write the equation in this form.

$$\frac{\sin[\alpha(\lambda-1)]}{\alpha(\lambda - 1)} = \frac{\sin(2\alpha)}{2\alpha}$$

Simply put, we're looking for $f(\alpha(\lambda-1))=f(2\alpha)$ where $f(x) = \dfrac{\sin x}{x}$. Everything from here on will involve the sinc function.

Here's a plot

If $\alpha \ne 0$, there are three solutions:, $\lambda = 1$ and $\lambda-1 = \pm 2$ or $\lambda = -1, 3$.

A special case occurs when $\sin (2\alpha) = 0$, or $\alpha = \dfrac{n\pi}{2}$, $n \ne 0$. Then you get rational solutions $\lambda = 1 + \dfrac{2m}{p}$, where $m$ is an integer and $p$ is any factor of $n$. This solution set is countably infinite.

EDIT: You don't want $\alpha = \dfrac{n\pi}{2}$ since this makes the original equation undefined, but everything below this still holds.

For any other case, you'll always get a finite number of solutions, since the sinc function decays in amplitude. Obtaining a definite count is possible, but rather complicated, since it requires knowing the extrema of the sinc function. You'll have to use numerical methods from here.

It's easy to count the solution set if $f'(2\alpha) = 0$. Then there are no solution $|\lambda - 1| > 2$.

• There are no additional solution if $2\alpha$ is the minimum point in $\pm(\pi,2\pi)$
• There are 2 additional solutions if $2\alpha$ is the maximum point in $\pm(2\pi,3\pi)$
• There are 4 additional solutions if $2\alpha$ is the minimum point in $\pm(3\pi,4\pi)$

and so on. There are a total of $2n+1$ solutions (including $\lambda = 1$) for $|2\alpha| \in \big(n\pi, (n+1)\pi\big)$

If $f'(2\alpha) \ne 0$, then there may exist some $\beta$ such that $f'(\beta) = f'(2\alpha)$ and $f(\beta) = f(2\alpha)$. If this happens, there are once again $2n+1$ solutions for $|\beta| \in \big(n\pi, (n+1)\pi\big)$

Finally, if $f(2\alpha)$ does not coincide with any extrema, then it has to occur between two maxima or minima. We can then find a $c$ such that $f'(c) = 0$ and $|f(2\alpha)-f(c)|$ is minimal. Then there are $2n-1$ solutions for $|c| \in \big(n\pi, (n+1)\pi\big)$

Tldr; the problem boils down to counting the solutions of $\dfrac{\sin x}{x} = a$

For something somewhat more empirical, consider $\alpha = \frac{\pi}{4}$. For this value, your derived equation reduces to

$$2\sin \frac{(\lambda-1)\pi}{4} = \lambda-1$$

The below graph of the LHS and RHS as a function of $\lambda$ shows that there are possible solutions at $-1$, $1$, and $3$, and nowhere else.

For $\lambda = -1$, we have

$$\frac{\sin -\alpha}{\sin \alpha} - \frac{\cos -\alpha}{\cos \alpha} = -2$$ $$-1 - 1 = -2$$

which works. For $\lambda = 1$, we have

$$\frac{\sin \alpha}{\sin \alpha} - \frac{\cos \alpha}{\cos \alpha} = 0$$ $$1-1 = 0$$

which again works. And finally, for $\lambda = 3$, we have

$$\frac{\sin 3\alpha}{\sin \alpha} - \frac{\cos 3\alpha}{\cos \alpha} = 2$$ $$2 \cos (2x) + 1 - [2 \cos (2x) - 1] = 2$$

which works once more.

By inspection $\lambda=3$ for all values of $\alpha$. $\lambda = 1,-1$ are also solutions by inspection.

• thanks for the solution, another question about that equation, what if i want to know the number of solutions for real lambda – Ranjeet Bahadur Feb 23 '18 at 17:14
• $\lambda = 1$ works – saulspatz Feb 23 '18 at 17:24