Find the values of the other five trigonometric functions for angle S.

a) cos S=1/3 and the terminal side of angle S is in Q1.
b) tan S=-10/7 and the terminal side of S is in Q2.
c) sin S=-5/14 and the terminal side of S is in Q3.
d) sec S=13/11 and the terminal side of angle S is in Q4.

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1 year ago

To find the values of the other trigonometric functions for angle S, we will use the given information and the definitions of the trigonometric functions.

a) Given cos S = 1/3 and the terminal side of angle S is in Q1.
Since cos S is positive in Q1, we know that S is an acute angle.
To find the values of the other trigonometric functions, we can use the following definitions:
- sin S = โˆš(1 - cos^2 S) = โˆš(1 - (1/3)^2) = โˆš(1 - 1/9) = โˆš(8/9) = โˆš8/3
- tan S = sin S / cos S = (โˆš8/3) / (1/3) = โˆš8
- csc S = 1 / sin S = 1 / (โˆš8/3) = 3/โˆš8 = 3โˆš8/8
- sec S = 1 / cos S = 1 / (1/3) = 3
- cot S = 1 / tan S = 1 / โˆš8 = โˆš8/8

b) Given tan S = -10/7 and the terminal side of S is in Q2.
Since tan S is negative in Q2, we know that S is an obtuse angle.
To find the values of the other trigonometric functions, we can use the following definitions:
- sin S = tan S / โˆš(1 + tan^2 S) = (-10/7) / โˆš(1 + (-10/7)^2) = (-10/7) / โˆš(1 + 100/49) = (-10/7) / โˆš(49/49 + 100/49) = (-10/7) / โˆš(149/49) = (-10/7) / (โˆš149/7) = -10/โˆš149
- cos S = 1 / โˆš(1 + tan^2 S) = 1 / โˆš(1 + (-10/7)^2) = 1 / โˆš(1 + 100/49) = 1 / โˆš(49/49 + 100/49) = 1 / โˆš(149/49) = 1 / (โˆš149/7) = 7/โˆš149
- csc S = 1 / sin S = 1 / (-10/โˆš149) = -โˆš149/10
- sec S = 1 / cos S = 1 / (7/โˆš149) = โˆš149/7
- cot S = 1 / tan S = 1 / (-10/7) = -7/10

c) Given sin S = -5/14 and the terminal side of S is in Q3.
Since sin S is negative in Q3, we know that S is an obtuse angle.
To find the values of the other trigonometric functions, we can use the following definitions:
- cos S = โˆš(1 - sin^2 S) = โˆš(1 - (-5/14)^2) = โˆš(1 - 25/196) = โˆš(196/196 - 25/196) = โˆš(171/196) = โˆš171/14
- tan S = sin S / cos S = (-5/14) / (โˆš171/14) = -5/โˆš171
- csc S = 1 / sin S = 1 / (-5/14) = -14/5
- sec S = 1 / cos S = 1 / (โˆš171/14) = 14/โˆš171
- cot S = 1 / tan S = 1 / (-5/โˆš171) = -โˆš171/5

d) Given sec S = 13/11 and the terminal side of S is in Q4.
Since sec S is positive in Q4, we know that S is an acute angle.
To find the values of the other trigonometric functions, we can use the following definitions:
- cos S = 1 / sec S = 1 / (13/11) = 11/13
- sin S = โˆš(1 - cos^2 S) = โˆš(1 - (11/13)^2) = โˆš(1 - 121/169) = โˆš(169/169 - 121/169) = โˆš(48/169) = โˆš48/13
- tan S = sin S / cos S = (โˆš48/13) / (11/13) = โˆš48/11
- csc S = 1 / sin S = 1 / (โˆš48/13) = 13/โˆš48 = 13โˆš48/48 = 13โˆš3/4
- cot S = 1 / tan S = 1 / (โˆš48/11) = 11/โˆš48 = 11โˆš48/48 = 11โˆš3/4

In summary, for each given case:
a) sin S = โˆš8/3, tan S = โˆš8, csc S = 3โˆš8/8, sec S = 3, cot S = โˆš8/8
b) sin S = -10/โˆš149, cos S = 7/โˆš149, csc S = -โˆš149/10, sec S = โˆš149/7, cot S = -7/10
c) cos S = โˆš171/14, tan S = -5/โˆš171, csc S = -14/5, sec S = 14/โˆš171, cot S = -โˆš171/5
d) sin S = โˆš48/13, cos S = 11/13, tan S = โˆš48/11, csc S = 13โˆš3/4, sec S = 13/11, cot S = 11โˆš3/4