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The question is to find the values of a real number $\lambda$ for which the following equation is satisfied for all real values of $\alpha$ which are not integral multiples of $\pi/2$ $${\sin\lambda\alpha\over \sin\alpha}-{\cos\lambda\alpha\over \cos\alpha}=\lambda-1$$

All I could do was to guess some values that just came to mind by observation, like $-1,1,3$

What should be a more mathematical way to find all possible values of $\lambda$?

SOURCE: KVPY 2015 SB stream

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  • $\begingroup$ KVPY 2015 is now over, btw. $\endgroup$ – najayaz Nov 6 '15 at 6:19
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Cross multiplying gives $$ \sin((\lambda - 1)\alpha) = \frac{(\lambda -1)}{2}\sin(2\alpha) \qquad (\ast) $$ Since both sides are continuously differentiable (as functions of $\alpha$), we obtain $$ \cos((\lambda - 1)\alpha) = \cos(2\alpha) $$ At $\alpha = \pi/4$, we get $$ \cos((\lambda - 1)\pi/4) = 0 $$ $$ \Rightarrow \lambda -1 \in \{2,-2,6,-6,10,-10, \ldots\} $$ Also $\lambda = 1$ is clearly a solution to $(\ast)$, so we get $$ \lambda \in \{1,3,7,11,\ldots\}\cup\{-1,-5,-9,\ldots\} $$

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  • $\begingroup$ Why did you put $\alpha=\pi/4$? $\endgroup$ – najayaz Nov 6 '15 at 7:11
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To validate as an identity for all real $\alpha$

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

we are allowed to take small $\alpha$ values $ \sin x \approx x $

A root as stationary point/ singular solution exists for all $\alpha$ is obtained as :

$ \lambda - 1=0 , \lambda = 1 $

For other solutions around this point we have to numerically find roots like for the roots of : $ \tan x = a \cdot x ,$ tangent curve cutting variable slope line.

Shall add to this soon.

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