Proving $\cos^{-1}(-x)=\pi-\cos^{-1}x$ without geometry.

Question

Let $$ \cos^{-1}x=a \implies x=\cos a $$ and $$ \cos^{-1}(-x)=b \implies -x=\cos b $$ Hence we have $$ \cos a+\cos b=0 $$ Using $$ \cos(A+B)+\cos(A-B)=2\cos A\cos B $$ with $$ A=\frac{a+b}{2} \\ \text{ and } \\ B=\frac{a-b}{2} $$ we get $$ \cos a+\cos b=2\cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right)=0 $$ with which $$ \cos\left(\frac{a+b}{2}\right)=0 \text{ or } \cos\left(\frac{a-b}{2}\right)=0\\ \implies a=\pi-b \text{ or } a=\pi+b $$

Now, to choose between the two, I'm making the below argument:
To find the inverse of a function, the function has to be one-to-one. Hence in our case, both $\cos a$ and $\cos b$ have to be one-to-one, which is possible only when $$ n\pi\leq a,b \leq (n+1)\pi \text{, }n\in\mathbb{Z} $$

From the above, we get, $$ -n\pi \leq \pi-b \leq (-n+1)\pi $$ and $$ (n+1)\pi \leq \pi+b \leq (n+2)\pi $$

The ranges can be reconciled for $a$ and $\pi-b$ by taking $n=0$ and we get $$ 0 \leq a \leq \pi \\ 0 \leq \pi-b \leq \pi $$

But no value of $n\in\mathbb{Z}$ can simultaneously reconcile the ranges of $a$ and $\pi+b$.

Therefore, we conclude that $$ \cos^{-1}(-x)=\pi-\cos^{-1}(x) $$

From this I also learnt that the inverse function is meaningful when the angle is discussed in the range $[0,\pi]$. Is this line of argument mathematically fool-proof?

Answer 1

That looks correct, but it's much simpler to say that, if $x\in[-1,1]$,\begin{align}\cos\bigl(\pi-\arccos(x)\bigr)&=-\cos\bigl(\arccos(x)\bigr)\\&=-x\\&=\cos\bigl(\arccos(-x)\bigr)\end{align}and that therefore, since $\pi-\arccos(x),\arccos(-x)\in[0,\pi]$ and since the restriction of $\cos$ to that interval is injective, $\pi-\arccos(x)=\arccos(-x)$.

Answer 2

If $0\leq{x}\leq\pi\implies$ we can use the next formula relatively of the both parts of the provided expression by you: $$ \bbox[lightgreen] { \cos(\arccos(x))=x. } \qquad\qquad\qquad\qquad{(1)} $$ Therefore $$ \cos(\arccos(-x))=\cos(\pi-\arccos(x)), $$ But another formula also exists: $$ \bbox[lightgreen] { \arccos(-x)=\pi-\arccos(x) } \qquad\qquad\qquad\qquad{(2)} $$ Applying $(2)$ to the previous expression we get: $$ \cos(\arccos(-x))=\cos(\arccos(-x)). $$ Good luck!

Answer 3

First of all let us prove that $\arccos(x)+\arccos(-x)$ is a constant. In fact, differentiate it we get $$ (\arccos(x)+\arccos(-x))'=-{\frac {1}{\sqrt {-{x}^{2}+1}}}+{\frac {1}{\sqrt {-{x}^{2}+1}}}=0. $$ Thus $\arccos(x)+\arccos(-x)=C$ for some $C$.

Now, to find $C$ just put $x=0.$ We obtain that

$\frac{\pi}{2}+\frac{\pi}{2}=C \implies C=\pi,$ as required.

Answer 4

Let $f(x) = \cos^{-1}(-x) + \cos^{-1}x$. Note that $$ f^{\prime}(x) = -\dfrac{1}{\sqrt{1 - (-x)^2}}\cdot (-1) - \dfrac{1}{\sqrt{1 - x^2}} = 0 \quad \Rightarrow \quad f(x) = k, \quad k \in \mathbb{R} $$ For $x = 0$, we have $k = \cos^{-1}(-0) + \cos^{-1}0 = \pi/2 + \pi/2 = \pi$. Thus, $\cos^{-1}(-x) = \pi - \cos^{-1} x$.

Answer 5

We have $$ \cos(\pi-\cos^{-1}x)=\cos\pi\cos\cos^{-1}+\sin\pi\sin\cos^{-1}x=-x $$ Since $0\le \pi-\cos^{-1}x\le \pi$, applying $\cos^{-1}$ on both sides of the identity $\cos(\pi-\cos^{-1}x)=-x,$ we get $$ \pi-\cos^{-1}x=\cos^{-1}(-x) $$

Answer 6

Another way:

Use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ replacing $y$ with $-x$ so that $xy=-x^2<0$

$$\arcsin(x)+\arcsin(-x)=\arcsin(0)=0$$

Now use Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$