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Unformatted text preview: 2 Solution: We have already seen that the restriction of the function sin to the interval /2, /2 is
strictly increasing and that the range of this function is the interval 1, 1 . The function arcsin is
therefore a strictly increasing continuous function from 1, 1 onto /2, /2 and, from the facts about
differentiation of inverse functions studied earlier we know that for every number u between 1 and 1 we
have
1
arcsin u
cos arcsin u
Since the function cos is nonnegative on the interval /2, /2 we deduce that
1
1
1
arcsin u
cos arcsin u
1 u2
1 sin 2 arcsin u
Therefore whenever 1 u 1 we have
arcsin u Þ0 u 1
1 t2 dt. 4. By analogy with the preceding exercise, give a definition of the function arctan and deduce that if u is any
real number then
u
1 dt
arctan u Þ
0 1 t2
The function tan is, of course, defined by the equation
sin
tan x cos x
x
whenever cos x 0. At any such number x we have
1
0
tan x
cos 2 x
,
is a strictly
and we conclude that the restriction of the function tan to the interval
22
increasing function whose derivative is everywhere positive. Furthermore, since
sin
lim tan x lim cos x Ý
x
x /2
x /2
and
x sin
lim tan x lim cos x Ý,
x
/2
x /2
the range of the restriction of tan to
,
is the entire set R.
22
,
. We deduce from the
We define arctan to be the inverse function of the restriction of tan to
22
theorem on differentiation of inverse functions that if x is any real number we have
1
1
1
1.
arctan x
1
tan arctan x
1 tan 2 arctan x
1 x2
2
cos arctan x It follows from the fundamental theorem that if u is any real number we have 379 arctan u Þ0 u 1 dt.
1 t2 5. Prove that the restriction of the function cos to the interval 0, is a strictly decreasing function from the
interval 0, onto the interval 1, 1 . Prove that if the function arccos is now defined to be the inverse
function of this restriction of cos then for every number u
1, 1 we have
Þu
1
arccos u
dt.
2
0
1 t2
We know that cos 0 1 and cos 1 and that
cos x sin x 0
whenever 0 x . Therefore cos is a strictly decreasing differentiable function from the interval
0, onto 1, 1 whose derivative is positive at every number between 1 and 1. We define arccos
to be the inverse function of the restriction of the function cos to 0, . Note that because sin u 0
whenever 0 u we have the identity
sin arccos x
whenever 1 x
1 x 1 then cos 2 arccos x 1 1 x2 1. We deduce from the theorem on differentiation of inverse functions that if
arccos x 1
cos arccos x 1
sin arccos x 1
1 x2 and it follows from the fundamental theorem that if 1 u 1 then
u
1
arccos u arccos 0 Þ
dt
0
1 t2
u
1
Þ
dt.
2
0
1 t2
6. Prove that if x and y are real numbers that are not both zero and if
arccos x2 x
y2 then
sin y . x2 y2 Deduce that if x and y are real numbers that are not both zero then there exists a positive number r and a real
number
0, 2 such that x r cos and y r sin .
Suppos...
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 Fall '08
 STAFF
 Math, Calculus

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