SOLUTION: Find the exact value of tan (2 θ) cos (θ) = - (15)/(17) and θ terminates in QII Select one: a. 161/289 b. 120/289 c. - (240)/(161) d. - (240)/(289)

Algebra ->  Test -> SOLUTION: Find the exact value of tan (2 θ) cos (θ) = - (15)/(17) and θ terminates in QII Select one: a. 161/289 b. 120/289 c. - (240)/(161) d. - (240)/(289)      Log On


   

Question 1093700: Find the exact value of tan (2 θ)
cos (θ) = - (15)/(17) and θ terminates in QII
Select one:
a. 161/289
b. 120/289
c. - (240)/(161)
d. - (240)/(289)
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
T stands for theta.

you have cos(T) = -12/17 and T terminates in the second quadrant.

17 is the hypotenuse of the right triangle formed and 15 is one leg of the right triangle formed.

the other leg of the right triangle formed is equal to sqrt(17^2 - 15^2) = 8.

this is found through use of the pythagorean formula.

the legs of the right triangle formed in the second quadrant have a value of x = -15, y = 8 and hypotenuse = 17.

you have tan(T) = 8/-15 which is equivalent to -8/15.

the formula for tan(2T) is tan(2T) = 2 * tan(T) / (1 - tan^2(T)

tan(T) = -8/15.

tan^2(T) = (-8/15)^2 = 64/225.

2 * tan(T) = -16/15.

1 - tan^2(T) = 161/225.

formula becomes tan(2T) = (-16/15) / (161/225) which becomes tan(2T) = -16/15 * 225/161 which becomes tan(2T) = (-16*225) / (15/161) which then becomes tan(2T)
= (-16*15) / 161 which finally becomes tan(2T) = -240/161.

that's selection c.

the angle T is equal to arccos(-15/17) = 151.9275131 degrees.

that's in the second quadrant.

the angle (2T) is equal to 2 times that = 303.8550261.

that's in the fourth quadrant.

tan(T) = tan(151.9275131) = -.53333333333 which is the same as -8/15.

tan(2T) = tan(303.8550261) = -1.49068323 which is the same as -240/161.

cos(T) = cos(151.9275131) = -.8823529412 which is the same as -15/17.

everything checks out so the solution is looking good.

selection c is your answer.