CHAPTER 27
222
27.12
27.13
27.14
27.15
27.16
27.17
27.18
27.19
27.20
27.21
27.22
27.23
By definition, the value 6 must be an angle in either the first or third quadrant. Since esc 6 = - V2, 0 is in
third quadrant, sin 0 = -1 /V5, and 0 = TT + -rr/4 = Sir/4.
Find sin 6.
Since
the first quadrant, the + sign applies and sin 6 = 2V2/3.
and sin0 = ±V1- cos 0 = ±
= ±2V2/3. Since 0 is in
In Problems 27.16-27.20, compute the indicated functional value.
= ±V15/4. But cos0>0, since
0 < 6 < IT 12 and sin 6 must be positive, sin 6 = VI - cos' 0 =
Since
0 < 0 < ir/2, and therefore, tan 0 >0. But tan 6 = Vsec 6 - 1 =
Let
and 3 are positive, 8, and 0 2 are in the first quadrant. Then
Let 6 l = cos
Find the domain and range of the function f(x) = cos (tan ' x).
Since the domain of tan ' x is the whole set £% of real numbers and cos u is defined for all u, the domain of/is
9?. The values of tan"
1 x form the interval (-ir/2, -rr/2), and the cosines of angles in that interval form the
interval (0,1], which is, therefore, the range off.
In Problems 27.22-27.25, differentiate the given function.
tan
l x + cot ' x.
sin ' x + cos ' x.
esc"
1 (-V2).
cor'(-l).
The value 0 is, by definition, in (0, TT). Since the argument is negative, 6 is in the second quadrant,
cot0 = -l, tan 0 = !/(-!) = -1, 0 = 7T/2+ -rrl4 = 3ir/4.
Let 0 = cos '
6 is in (0, 7r/2). cos 0 =
Let 0 = sin '
Find cos 0.
Since
2 0 = ±
-77/2<0<0. Hence, cos0 = V15/4.
sin (
Let 0 = cos '
Since
tan (sec '
cos (sin
Let 0 = sec
l
and 0 2 = sec ' 3. Since
sin (cos
cos
cos (6, + 0 2 ) = cos 0. cos 0, - sin 0, sin 0,
and 0, = tan 2. Then 0, and 0, are in the first quadrant, sin
sin (0j - 0 2 ) = sin 0 t cos 0 2 - cos 0, sin 0 2 = (2V6/5)(1 /V5)
;2/V5) = (2V6-2)/5V5.
sin ' (sin TT).
sin TT = 0. Hence, sin" (sin TT) = sin 0 = 0. Note that sin (sin x) is not necessarily equal to x.
222
27.12
27.13
27.14
27.15
27.16
27.17
27.18
27.19
27.20
27.21
27.22
27.23
By definition, the value 6 must be an angle in either the first or third quadrant. Since esc 6 = - V2, 0 is in
third quadrant, sin 0 = -1 /V5, and 0 = TT + -rr/4 = Sir/4.
Find sin 6.
Since
the first quadrant, the + sign applies and sin 6 = 2V2/3.
and sin0 = ±V1- cos 0 = ±
= ±2V2/3. Since 0 is in
In Problems 27.16-27.20, compute the indicated functional value.
= ±V15/4. But cos0>0, since
0 < 6 < IT 12 and sin 6 must be positive, sin 6 = VI - cos' 0 =
Since
0 < 0 < ir/2, and therefore, tan 0 >0. But tan 6 = Vsec 6 - 1 =
Let
and 3 are positive, 8, and 0 2 are in the first quadrant. Then
Let 6 l = cos
Find the domain and range of the function f(x) = cos (tan ' x).
Since the domain of tan ' x is the whole set £% of real numbers and cos u is defined for all u, the domain of/is
9?. The values of tan"
1 x form the interval (-ir/2, -rr/2), and the cosines of angles in that interval form the
interval (0,1], which is, therefore, the range off.
In Problems 27.22-27.25, differentiate the given function.
tan
l x + cot ' x.
sin ' x + cos ' x.
esc"
1 (-V2).
cor'(-l).
The value 0 is, by definition, in (0, TT). Since the argument is negative, 6 is in the second quadrant,
cot0 = -l, tan 0 = !/(-!) = -1, 0 = 7T/2+ -rrl4 = 3ir/4.
Let 0 = cos '
6 is in (0, 7r/2). cos 0 =
Let 0 = sin '
Find cos 0.
Since
-77/2<0<0. Hence, cos0 = V15/4.
sin (
Let 0 = cos '
Since
tan (sec '
cos (sin
Let 0 = sec
l
and 0 2 = sec ' 3. Since
sin (cos
cos
cos (6, + 0 2 ) = cos 0. cos 0, - sin 0, sin 0,
and 0, = tan 2. Then 0, and 0, are in the first quadrant, sin
sin (0j - 0 2 ) = sin 0 t cos 0 2 - cos 0, sin 0 2 = (2V6/5)(1 /V5)
;2/V5) = (2V6-2)/5V5.
sin ' (sin TT).
sin TT = 0. Hence, sin" (sin TT) = sin 0 = 0. Note that sin (sin x) is not necessarily equal to x.