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In-Depth Information
(iii)
A
is monotonic decreasing and so a bijection. Domain is given.
The range is bounded by the values of the function at the
extremities of the domain.
51
A
(1)
0, so cos 0
1 and sin 0
0.
A
(
1)
, so cos
1 and sin
0.
A
(0)
A
(
1)
, so cos
0 and sin
1.
52
A
(cos
x
)
x
and cos is differentiable by 8.42, inverse functions.
By qn 8.40,
A
(cos
x
) · cos
x
1, so (
1/
(1
cos
x
)) · cos
x
1.
cos
x
sin
x
.
x
(
2 cos
x
)(
sin
x
)
cos
x
· sin
x
sin
x
sin
x
x
), so sin
x
)
(1
cos
.
(1
cos
53
L
sup
f
(
x
)
a
x
b
.
54 Thefunction cos is continuous on [0,
] and differentiable on (0,
)so
we may apply the Mean Value Theorem and get
cos
x
cos
c
sin
c
for some
c
,
x
c
cos
.
x
As
x
,
c
sin
c
,so
0 and cos
0.
x
So wehavecos
sin
x
on (0, 2
).
Similarly sin
cos
1.
56
f
(
x
)
cos(
a
x
) cos
x
sin(
a
x
) sin
x
sin(
a
x
) sin
x
cos(
a
x
) cos
x
0.
From qn 9.17,
f
is constant. But
f
(0)
sin
a
. Now put
x
y
a
to obtain
sin(
x
y
)
sin
x
· cos
y
cos
x
· sin
y
.
58 Apply qns 56 and 57 putting
y
x
, and usesin
x
x
cos
1.
cos
x
·sin
x
sin
x
· cos
x
cos
x
sin
x
59
tan
x
x
.
1
tan
x
x
cos
cos
Thetan function is continuous on (
) and has positivedrivativeand
is therefore strictly increasing, and so a bijection.
As
x
,
, cos
x
0, so 1/cos
x
; also sin
x
1, so tan
x
.
Similarly as
x
, tan
x
.
60 arctan is differentiable by qn 8.42, inverse functions.
1
arctan(tan
x
)
x
arctan
(tan
x
) tan
x
1
arctan
(tan
x
)
1
tan
x
