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$\cos\theta = \cos\phi$ and $\sin\theta = \sin\phi$ give $\theta = 2n\pi + \phi$ ($n$ being any integer)

But dividing 2nd equation by 1st one gives $\tan\theta = \tan\phi$ which in turn reduces to $\theta = n\pi + \phi$ ($n$ being any integer).

Where am I wrong?

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2 Answers

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Think it out.

$\sin(\theta + \pi) = -\sin(\theta)$

$\cos(\theta + \pi) = -\cos(\theta)$

So $\tan(\theta + \pi) = \frac {\sin(\theta + \pi)}{\cos(\theta + \pi) } = \frac {-\sin \theta}{-\cos \theta} = \frac {\sin \theta}{\cos \theta} = \tan {\theta}$.

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So $\tan {\theta} = \tan {\phi}$

Means either $\sin{\theta} = \sin{\phi}$ and $\cos{\theta} = \cos{\phi}$ OR $\sin{\theta} = -\sin{\phi}$ and $\cos{\theta} = -\cos{\phi}$.

Which means $\theta = 2k\pi + \phi$ OR $\theta = (2k+1)\pi + \phi$.

Which means $\theta = n*\pi + \phi$.

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If you know nothing about the functions $f(x)$ and $g(x)$ except that they have period $l$ then you could deduce that $f(x) + g(x)$ and also multiplication and division of the two have period $l$ at most but you could not rule out a shorter period. The period might be a submultiple of $l$.

In this case, you might say that $sin$ and $cos$ are "anti-periodic" with anti-period $\pi$. In general, my "anti-periodic" would mean: $f(x + l) = -f(x)$. Now, it should be obvious that multiplying or dividing two anti-periodic functions gives you a regular periodic one. Also, it is obvious that if $f$ has anti-period $l$ then it has regular period $2l$ which might explain why it is not a common term.

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