082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

146 Chapter 3 The trigonometric functions
cos t
1
0.3685
A
B
p
D
C 2p t
–0.3685
0 p –2 3p –2
Figure3.21
–1
A andDare solution pointsfor
cost = 0.3685.Band Care
solutionpointsforcost = −0.3685.
This is the solution between 0 and π , that is at point A. Using the symmetry of
2
Figure 3.21 the other solution atpoint Dis
t = 2π −1.1934 = 5.0898
The required solutions aret = 1.1934 and 5.0898.
(b) The graph in Figure 3.21 shows there are two solutions of cost = −0.3685. These
solutions areatpoints Band C. Given
cost = −0.3685
then usingascientific calculator we have
t = cos −1 (−0.3685) = 1.9482
This isthe solution given by pointB. By symmetrythe other solution atpointCis
t = 2π −1.9482 = 4.3350
The required solutions aret = 1.9482 and 4.3350.
Example3.17 Solve
(a) tant = 1.3100for0 t 2π
(b) tant = −1.3100 for0 t 2π
Solution Figure3.22showsagraphofy = tant fort = 0tot = 2πtogetherwithhorizontallines
y = 1.3100 andy = −1.3100.
(a) Thereisasolutionoftant = 1.3100between0and π andasolutionbetween πand
2
3π
. These aregiven by points Aand C.
2
tant = 1.3100
t = tan −1 (1.3100) = 0.9188