How to convert arccos to arctan?
Solution 1
Consider the point $(A,B) \ne (0,0)$ corresponding to the angle $\theta$
Then $\tan \theta = \dfrac{B}{A}$ and $\cos \theta = \dfrac{A}{\sqrt{A^2+B^2}}$
We can tentatively conclude that
$\arccos\frac{A}{\sqrt{A^2+B^2}}=\arctan\frac{B}{A} = \theta$
but we need to check the ranges first.
The range of $\arccos$ is the interval $\left[0, \pi \right]$.
The range of $\arctan$ is the interval $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]$.
The intersection of these two ranges is the interval $\left[0, \dfrac{\pi}{2} \right]$.
So the formula is only true when $A \gt 0, B \ge 0$.
Solution 2
No. For $\arccos$ maps to $[0,\pi]$ and $\arctan$ maps to $(-\frac\pi2,\frac\pi2)$, so any $(A,B)$ leading to negative $\arctan$ or to and $\arccos$ that is $\ge \frac\pi2$, will fail. Also, multiplying both (non-zero) $A$ and $B$ by $-1$ does not change $\frac BA$, but negates $\frac A{\sqrt{A^2+B^2}}$ thus replaces the left hand side by $\pi$ minus its original value.
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Updated on May 01, 2020Comments
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GambitSquared over 3 years
Is this true? $$\arccos\frac{A}{\sqrt{A^2+B^2}}=\arctan\frac{B}{A}$$ If so, how can one show it?
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Damian Reding over 7 yearsMaybe try using $1+\tan^2 (x)=\frac{1}{\cos^2(x)}$
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Admin over 7 yearsWhat do you know about A, B?
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laugh salutes Monica C over 7 yearsYou should add "$+n\pi$" (for some integer n) to one side.
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