Question

Would anyone please help me find the value of these two expressions without using a calculator?
cos(arccos 1 + arcsin 0)
tan[arccos(-1)-arccos 1)]

Answer 1

\[\cos \left( \cos^{-1} 1+\sin^{-1} 0 \right)=\cos \left( 0+0 \right)=\cos 0=1\]

Answer 2

Oh... wait but the inverse...
the inverse confuses me...

Answer 3

so you just have to apply the unit circle?
kind of like pretending inverse doesn't exist there?

Answer 4

\[\tan [\cos^{-1} (-1)-\cos^{-1} 1]=\tan [ \pi-0]=\tan \pi=0\]

Answer 5

\[\arccos(x)=\int\limits_{}^{} \frac{ - 1 }{ \sqrt{1-x^{2}} }\]
let x=1

\[\arcsin(x)=\int\limits_{}^{} \frac{ 1 }{ \sqrt{1-x^{2}} }\]
let x=0

then do cos of the sum

Answer 6

Oh okay! Thank you!

Answer 7

I think I understand now.