by Andrey Kudinov
Last Updated September 14, 2018 15:20 PM

I was solving an apperently simple equation but then it turned out I missed one of the solutions that somedoby else find using another method.

$\cos x=\sin x$ was the problem.

I tried the associated angle method,so $\sin(x)=\cos(\frac{\pi}{2} - x)$. Performing the substitution I could equal the argument of Cos and solve the equation in $x$. It turned out I missed $5/4 \pi$ and I cannot understand why. Can somebody explain me the right way to using this method?

So, we have $\dfrac\pi2-x=2n\pi\pm x$ where $n$ is any integer

$$\frac{\pi}{2} -x = x+ k2\pi$$ $$ -2x= -\frac{\pi}{2} +k2\pi$$ $$x= \frac{\pi}{4} - k\pi$$ Which for any integer $k$ , would give the correct result.

Indeed, $\sin(x) = \cos(\frac\pi2 -x )$, therefore $\sin(x)=\cos(x)$ is equivalent to $\cos(\frac\pi2 -x ) = \cos(x)$.

However, note that trigonometric functions are periodic with a period of $2\pi$ and that cosine is an even function, i.e. $$\cos(x) = \cos(x + 2\pi k), \quad k \in \mathbb Z$$ and $$\cos(x) = \cos (-x),$$ or together $$\cos(x) = \cos(\pm x + 2\pi k), \quad k \in \mathbb Z.$$

Using this, $\sin(x)=\cos(x)$ becomes $$\frac\pi2 - x = \pm x + 2\pi k, \quad k \in \mathbb Z.$$

After rearranging, this finally becomes $$x = \left(n + \frac14\right)\pi, \quad k \in \mathbb Z.$$

Assume $\sin x = \cos x$.

If $\cos x = 0$ then $\sin^2x + \cos^2 x = 1 $ implies $\sin x = \pm 1$, which is a contradiction.

Therefore certainly $\cos x \ne 0$ so $\sin x = \cos x$ implies $\tan x = \frac{\sin x}{\cos x} = 1$. Therefore $$x = \frac{\pi}4 + k\pi, \quad k \in \mathbb{Z}$$

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