A trigonometric equality

Question

Can you help me prove: $\arccos \frac{y}{\sqrt{y^2 + x}} = \mathrm{arccot} \frac{y}{\sqrt{x}}$?

I could solve this problem myself, but maybe you can show me a simple way to prove this and similar equalities?

And no, it is not a homework.

Answer 1

Assuming $x>0$ we have $\left|\dfrac y{\sqrt{y^2+x}}\right|\le\dfrac{|y|}{\sqrt{y^2}}=1$, so $$\cos^2\left(\arccos\frac y{\sqrt{y^2+x}}\right)=\frac{y^2}{y^2+x}$$ and $$\cos^2\left(\text{arccot}\frac y{\sqrt x}\right)=\frac{\cot^2\left(\text{arccot}\frac y{\sqrt x}\right)}{1+\cot^2\left(\text{arccot}\frac y{\sqrt x}\right)}=\frac{\frac{y^2}x}{1+\frac{y^2}x}=\frac{y^2}{y^2+x}$$

Note that the range of $\arccos$ and $\text{arccot}$ is within $[0,\pi]$.

The function $\cos$ is injective on that interval, so it suffices to show that $\arccos\frac y{\sqrt{y^2+x}}>\frac\pi2$ only if $\text{arccot}\frac y{\sqrt x}>\frac\pi2$, but that's easy to see as both happen only when $y<0$.

Answer 2

We will require $x > 0$.

The $\arccos$ is an angle in between $0$ and $\pi$. Let $\theta = \arccos \dfrac{y}{\sqrt{y^2 + x}}$ so that $\cos \theta = \dfrac{y}{\sqrt{y^2 + x}}$.

Assume for simplicity that $y > 0$. Then $0 < \cos \theta < 1$ so that $0 < \theta < \dfrac \pi 2$. Draw a right triangle with one acute angle $\theta$. You can label the adjacent side with length $y$ and the hypotenuse with length $\sqrt{y^2 + x}$. The Pythagorean theorem implies that the side opposite $\theta$ has length $\sqrt x$. It follows that $$\cot \theta = \frac{y}{\sqrt{x}}$$ using the rule "adjecent over opposite". This means (since $\theta$ is in quadrant I) that $$\theta = \mathrm{arccot} \frac{y}{\sqrt{x}}.$$

The same idea works if $y \le 0$, but the diagram will need to be modified accordingly.

Answer 3

Let $\displaystyle\mathrm{arccot}\dfrac y{\sqrt x}=\theta$

$\displaystyle\implies (i)\cot\theta=\dfrac y{\sqrt x}$ and using this,

$\displaystyle(ii) 0<\theta<\pi\implies\sin\theta>0$ and consequently $\displaystyle\sin\theta=+\frac1{\sqrt{1+\cot^2\theta}}=+\sqrt{\frac x{y^2+x}}$

As $\displaystyle\cot\theta=\frac{\cos\theta}{\sin\theta},\cos\theta= \cot\theta\cdot\sin\theta=\frac y{\sqrt{y^2+x}}$

Again if $\displaystyle\arccos\frac y{\sqrt{y^2 + x}}=\phi$ using this, $\displaystyle0\le\phi\le\pi$

So, $\displaystyle\mathrm{arccot}\dfrac y{\sqrt x}=\theta=\arccos\frac y{\sqrt{y^2 + x}}$ if $\displaystyle0<\theta<\pi$

Answer 4

Let $$ z=\arccos \frac{y}{\sqrt{y^2 + x}} $$ then $$ \cos z=\frac{y}{\sqrt{y^2 + x}}\qquad\rightarrow\qquad\sin z=\frac{\sqrt{x}}{\sqrt{y^2 + x}} $$ and $$ \cot z=\frac{\cos z}{\sin z}=\frac{\frac{y}{\sqrt{y^2 + x}}}{\frac{\sqrt{x}}{\sqrt{y^2 + x}}}=\frac{y}{\sqrt{x}}\qquad\rightarrow\qquad z=\text{arc}\cot\frac{y}{\sqrt{x}}\qquad\text{iff}\qquad0< z<\pi\qquad\qquad\blacksquare $$