sin(x+ h)− sin(x) sin(h)cos(x )+ cos(h)sin(x)− sin (x)
--------h-------- = ----------------h---------------
sin (h ) 1− cos(h)
= cos(x)--h---− sin (x) ---h-----
sin (h ) 1− cos(h)sin(h)
= cos(x)------− sin (x) ---------------
h sin(h) h
From what was shown above 5.5, 5.6, it follows that the second term on the last line
converges to 0 as h → 0 and the first term converges to cos
= y as the above. Now
applying the quotient rule to find tan^{′}
(x )
,
cos2 (x) − (− sin (x ))sin(x) 1
tan′(x) = -----------2-----------= --2----= sec2(x) = 1 + tan2 (x)
cos (x) cos (x)
the last being a well known identity which says essentially that cos^{2}
(x)
+ sin^{2}
(x)
= 1.
Then as before, y = tan
(x )
,
( 2 ) dx ( 2) dx
1 = 1 + tan (x ) dy-= 1 + y dy-
and so
′ --1---
arctan (y) = 1 +y2
You can do all the other trigonometric functions and their inverses the same way.
Of course none of them have inverses unless their domains are restricted as
above. For example, x ∈
[− π, π ]
2 2
in order that sin will be one to one. The
choice of interval on which the function is one to one is somewhat arbitrary. One
could have done the same thing for arcsin if the interval had been
[3π, 5π]
2 2
instead of
[− π, π]
2 2
for example. However, it is traditional to pick the interval to
which the function is restricted to be that interval closest to 0 such that the
function is one to one and maps onto its maximum range. The following table
sumarizes the derivatives of the trigonometric functions and their inverses. In
the table D will be the domain of the function and R will be the range of the
function.