Finding the exact Trig function

Question

Find the exact value of the trigonometric function given that sin u = -12/13 and cos v = -21/29. (Both u and v are in Quadrant III.)

tan(u − v)

So what I got was Tanu=-12/5 TanV=-20/21

and therefore, (-12/5)-(-20/21)/1+(-12/5)(-20/21) = -152/345 BUT apparently it is wrong? Does anyone have any input on this?

Answer 1

You just got the sign wrong. It's positive. Tangent values in Q3 are positive.

The angle subtraction identity for tangent is:

Obviously we need to know what tan(u) and tan(v) are.

Since sin(u) = opp/hyp = -12/13 then the opposite side is -12 and the hypotenuse is 13. Using the Pythagorean Theorem and the fact that the angle is in Q3, the adjacent side is -5 meaning tan(u) = -12/-5 or tan(u) = 12/5

Since cos(v) = adj/hyp = -21/29 then the adjacent side is -21 and the hypotenuse is 29. Using the Pythagorean Theorem and the fact that the angle is in Q3, the opposite side is -20 meaning tan(v) = -20/-21 or tan(v) = 20/21

So tan(u - v) = (12/5 - 20/21)/(1 + 12/5•20/21)

In the numerator:

12/5 = 252/105

20/21 = 100/105

So (12/5 - 20/21) = (252/105 - 100/105) = 152/105

In the denominator:

(1 + 12/5•20/21) = (1 + 16/7) = (7/7 + 16/7) = 23/7

So tan(u - v) = (12/5 - 20/21)/(1 + 12/5•20/21) = (152/105)/(23/7) = 152/345

Answer 2

Hi!

You are off to a good start.

The first thing I notice is your problem stipulated both u and v are in Quad III, so tan is positive there.

So tanu = 12/5 and tan v = 20/21.

So then tan(u - v) = (tan u - tan v) / (1 + tanu tan v) = (12/5-20/21)/(1+(12/5)(20/21)). Try calculating again with the sign change! :)