010ill 77 1417 PM 140 points STrig2 43042 My Notes Ask Your

010\'ill 77% 14:17 PM -1.40 points STrig2 4.3.042 My Notes Ask Your T Findm from the given information. cotx:2, 180\'

Solution

1) cot x = 2
tan x = 1/2
since x is in the third quadrant,
sin x = -1/5, cos x = -2/5

x/2 is in the second quadrant...
tan(x/2) = sin x / (1 + cos x) = -1/(5 - 2) = -2 - 5
sec^2(x/2) = 1 + tan^2(x/2) = 1 + 9 + 45 = 10 + 45

tan(x/2) = -2 - 5
cos(x/2) = -1/(10 + 45)
sin(x/2) = (2 + 5) / (10 + 45)

2) Let tan^-1 (12/5) = A
=> tanA = 12/5
=> cos(2A)
= (1 - tan^2 A)/(1 + tan^2 A)
= [1 - (12/5)^2] / [1 + (12/5)^2]
= (25 - 144) / (25 + 144)
= - 119/169
=> cos [2tan^-1 (12/5)] = - 119/169.

3)

sin = 1/5

cos = -sqrt(24)/5

sin2 = 2*sin *cos = 2*(1/5)*(-sqrt(24)/5) = -2*sqrt(24)/25

 010\'ill 77% 14:17 PM -1.40 points STrig2 4.3.042 My Notes Ask Your T Findm from the given information. cotx:2, 180\'Solution1) cot x = 2 tan x = 1/2 since x i

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