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I have a formula:

$\dfrac{a_1\cdot\sin(-k\cdot x + \phi_1)+a_2\cdot\sin(k\cdot x + \phi_2)}{\ a_1\cdot\cos(-k\cdot x + \phi_1)+a_2\cdot\cos(k\cdot x + \phi_2)} $

I have to prove that, when $ a_1 \neq a_2 $ , the formula depends on x.

If $ a_1 = a_2 $ , it is clear that the result is $ \tan\left(\dfrac{\phi_1+\phi_2}{2}\right) $, the result does not contains $x$.

Any suggestion?

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Hint

Give two different values to $x$ and notice how the result is different. For example take $x = \phi_1 / k$, $x = \phi_2 / k$

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  • $\begingroup$ Thanks, but $ x = -\dfrac{\phi_2}{k} $ is better for me, but anyway, the hint was helpful :) $\endgroup$ – User20141219 Jan 14 '15 at 9:40

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