inverse trig fcns: sin^(-1)(cos(tan^(-1)(1))) - ....

[sigma]

I need help with evaluating the followning function. I know that tan inverse is a 45 degree angle (or pi/4) but I'm not so sure what sin inverse of 1 is. I just want to know what the trick is when working with inverse trig functions.

\(\displaystyle \
\L\
\sin ^{ - 1} (\cos (\tan ^{ - 1} (1))) - \tan ^{ - 1} (\cos (\sin ^{ - 1} (1)))
\\)

 

[galactus]

Since you know \(\displaystyle \L\\tan^{-1}(1)=\frac{\pi}{4}\), then you should also know that \(\displaystyle \L\\sin^{-1}(1)=\frac{\pi}{2}\)

 

[soroban]

Hello, sigma!

I'll get you started . . .
. . And I'll assume we want the smallest positive angles.

\(\displaystyle \L \sin^{-1}\left(\cos\left[\tan^{-1} (1)\right]\right) \,- \,\tan^{-1}\left(\cos\left[\sin^{-1}(1)\right]\right)\)

The first term is: \(\displaystyle \L\:\sin^{-1}\left(\cos\underbrace{\left[\tan^{-1}(1)\right]}_{\downarrow}\right)\)
. . . . . . . . . . . . . . \(\displaystyle \L= \;\sin^{-1}\underbrace{\left(\cos\left[\frac{\pi}{4}\right]\right)}_{\downarrow}\)
. . . . . . . . . . . . . . . . \(\displaystyle \L=\;\underbrace{\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)}_{\downarrow}\)

. . . . . . . . . . . . . . . . . . . .\(\displaystyle \L=\;\;\frac{\pi}{4}\)


Now you work on the second term . . .