Show that $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$ for all $-1\le x \le 1$ (without using calculus!)

 

Show that $\cos^{-1}\cos x=\sin^{-1}\sin \left(x-\dfrac{\pi}{2}\right)+\dfrac{\pi}{2}$

There are two ways to see this result.

Graphical 'proof'

One is sketch the graphs and as they are both zigzags it is easy to see the horizontal and vertical translations through $\dfrac{\pi}{2}$, see diagrams

Algebraic proof

Expand,

$\begin{array}{rl}\sin^{-1}\sin \left(x-\dfrac{\pi}{2}\right) &= \sin^{-1}(\sin x \cos\dfrac{\pi}{2}-\cos x\sin\dfrac{\pi}{2})\\&=\sin^{-1}(-\cos x)\\&=-\sin^{-1}\cos x\end{array}$

But we know that $$\sin^{-1}p+\cos^{-1}p=\dfrac{\pi}{2}$$ for all $p$.

$\therefore \cos^{-1}\cos x+\sin^{-1}\cos x=\dfrac{\pi}{2}$

$\therefore \sin^{-1}\sin \left(x-\dfrac{\pi}{2}\right) = \cos^{-1}\cos x -\dfrac{\pi}{2}$

$\therefore \cos^{-1}\cos x = \sin^{-1}\sin \left(x-\dfrac{\pi}{2}\right) +\dfrac{\pi}{2}$

Graphs of compositions of inverse trig functions - $y=\sin^{-1}\cos x, y=\cos^{-1}\sin x,y=\sin\sin^{-1} x,y=\cos\cos^{-1} xy=\cos\sin^{-1} x$

 The graphs of these are either easy or based on previous graphs done (by translation left or right)

$y=\sin^{-1}\cos x$, 

(For $y=\sin^{-1}\sin x$ see this post   https://ripplemath.blogspot.com/2023/05/sketching-function-ysin-1sin-x-for.html)

$y=\cos^{-1}\sin x$,

(For $y=\cos^{-1}\cos x$ see this post https://ripplemath.blogspot.com/2023/05/sketching-function-ycos-1cosx-for.html )

$y=\sin\sin^{-1} x$,

$y=\cos\cos^{-1} x$

$y=\cos\sin^{-1} x$

Sketching the function $y=\cos^{-1}\cos(x)$ for $x\in\mathbb{R}$

 YouTube Video Link:  YOUTUBE CHANNEL

Question: Sketch $y=\cos^{-1}\cos x$ for $x\in\mathbb{R}$.

Basically you work it out for $0 \le x \le \pi $ first, which is fairly easy just using the definition of the inverse cosine function. Then you can use symmetry to guess the rest. 

Below we did work out the details for  $\pi \le x \le 2\pi$ and since the function has period $2\pi$ we know the rest of the graph.

The YouTube video talks through the basic step.

Below is a screen shot from the YouTube video for this problem.

Sketching the function $y=\sin^{-1}\sin x$ for $x\in\mathbb{R}$

 YouTube Video Link:  YOUTUBE CHANNEL

Question: Sketch $y=\sin^{-1}\sin x$ for $x\in\mathbb{R}$.

Basically you work it out for $-\pi/2 \le x \le \pi/2$ first, which is fairly easy just using the definition of the inverse function. Then you can use symmetry to guess the rest. Below we did work out the details for  $\pi/2 \le x \le 3\pi/2$ and since the function has period $2\pi$ we know the rest of the graph.

The YouTube video talks through the basic step.

Below is a screen shot from the YouTube video for this problem.

Finding Limits in a Nutshell - Youtube screenshots + link

Hi everyone, I have done a  quick YouTube video on Finding Limits in a Nutshell, here is the link on my youtube channel, Youtube.