Inverse Functions And Logarithms 3

Published On :2021-10-18 04:30:00

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Inverse Trigonometric Functions



We know that the sine function y=sin x is not one-to-one (use the Horizontal Line Test). But the function f(x)=sin x, frac{{ - pi }}{2} le x le frac{pi }{2} is one-to-one so we can find the inverse of it

[{sin ^{ - 1}}y = x Leftrightarrow y = sin x,frac{{ - pi }}{2} leqslant x leqslant frac{pi }{2}]
[Thus, - 1 leqslant y leqslant 1]


[Note:{sin ^{ - 1}}x ne frac{1}{{sin x}}]


Remember:


So now we know that the range of sine is between -1 and 1 
and the range of inverse sine is between frac{{ - pi }}{2} le x le frac{pi }{2}
for that, we should know the facts that
  1. [{sin ^{ - 1}}(sin x) = x,,for,frac{{ - pi }}{2} leqslant x leqslant frac{pi }{2}]
  2. [sin ({sin ^{ - 1}}x) = x,,for, - 1 leqslant x leqslant 1]





the same thing we can do with cos and tan 
[{cos ^{ - 1}}(cos x) = x,,for,0 leqslant x leqslant pi ]
[cos ({cos ^{ - 1}}x) = x,,for, - 1 leqslant x leqslant 1]




[{tan ^{ - 1}}x = y Leftrightarrow tan y = x,,,,and,,frac{{ - pi }}{2} leqslant y leqslant frac{pi }{2}]