The function $$f(x)=\operatorname{tg}\frac{11x}{34}+\operatorname{ctg}\frac{13x}{54}$$ has a period of $$918\pi$$, which is established as the smallest positive period. The tangent function $$\tan\frac{11x}{34}$$ has a repeating interval of $$\frac{34\pi}{11}$$, while the cotangent function $$\cot\frac{13x}{54}$$ repeats every $$\frac{54\pi}{13}$$. The least common multiple of these two intervals is $$918\pi$$, and the function does not exhibit any smaller repeating intervals due to the points of discontinuity occurring only at multiples of $$918\pi$$.
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The tangent and cotangent functions both have period $\pi$. So the function $\tan\frac{11x}{34}$ will repeat at intervals $\frac{34\pi}{11}$, and $\cot\frac{13x}{54}$ will repeat at intervals $\frac{54\pi}{13}$. You need to find the least common multiple of those two intervals.Andrei said:One period of the function $$f(x)=\operatorname{tg}\frac{11x}{34}+\operatorname{ctg}\frac{13x}{54}$$ is $$918\pi.$$ Please help me to prove that this is the smallest positive period. I can not use the most of trigonometric identities.
$$T=918\pi$$ is the least common multiple of those periods. I found this. But why it is the smallest positive period of $$f$$?Opalg said:You need to find the least common multiple of those two intervals.
You could look at the set of points at which $f(x)$ becomes infinite (or is undefined). That will happen at the points $x = \bigl(\frac{34}{11}k + \frac12\bigr)\pi$ and $x = \frac{54}{13}k\pi$ (and nowhere else). I think you should find that that set of points does not repeat at intervals of less than $918\pi$.Andrei said:$$T=918\pi$$ is the least common multiple of those periods. I found this. But why it is the smallest positive period of $$f$$?
For example, I consider the functions $$f_1(x)=\sin x$$ and $$f_2(x)=\operatorname{tg} x-\sin x$$, which both have $$2\pi$$ as main period. But then $$\pi$$ is the main period of $$f_1+f_2$$, which is not the least common multiple of $$2\pi.$$