Fig. 27-4
INVERSE TRIGONOMETRIC FUNCTIONS
221
27.3
27.4
27.5
27.6
27.7
27.8
27.9
27.10
27.11
The value 0 must be in the third quadrant. Since sec0 = -2V3/3, cos 0 = -3/2V3 = -V3/2. Thus
(see *'?. 27-4), e = it + ir/6 = 77T/6.
By definition, the value of sec ' x is either an angle in the first quadrant (for positive arguments) or an angle in
the third quadrant (for negative arguments). In this case, the value is in the first quadrant, so sec"
1 V2 =
cos~
1 l/V2=7r/4.
tan
1 1 is the angle 0 between - IT 12 and it 12 for which tan 0 = 1, that is, 0=ir/4.
sin (-V2/2) is the angle 0 between -ir/2 and ir/2 for which sin 0 = -V2/2. Clearly, 0 = -w/4.
sin
l V2/2 is the angle 0 between-77/2 and ir/2 for which sin0 = V2/2. Clearly, 0=ir/4.
Fig. 27-3
7T/6 is the angle 0 between-ir/2 and 7T/2 for which tan0 = (V3/3). So tan (V3/3) = 77/6.
Draw the graph of y = tan * x.
As Jt varies from-oo to+<», tan ' x varies from - ir/2 to 7r/2. The graph of >> = tan x is obtained from
that of .y = tanx [Fig. 27-2(a)] by reflection in the line y = x. See Fig. 27-2(6).
Show that Detail"
1 x) = !/(! + x
2 ).
In Problems 27.5-27.13, find the indicated number.
cos
1 (-V5/2) is the angle 0 between 0 and ir for which cos0 = -V3/2. It is seen from Fig. 27-3 that 0 is
the supplement of ir/6, that is, 0 = 5ir/6.
Let y = tan l x. Then tan .>> = *. By implicit differentiation, sec y-Dxy = l, D,y = I/sec y =
l/(l + tan
2 y) = l/(l + x
2 ).
cos"
1 (-V3/2).
sin"
1 (V2/2).
sin"
1 (-V2/2).
tan'
1 1.
tan"
1 (V3/3).
sec 'V2.
sec'
1 (-2V5/3).