Question

How do you find the value for `sin2theta`, `cos2theta`, and `tan2theta` and the quadrant in which `2theta` lies given `tantheta=-5/12` and `theta` is in quadrant II?

Answer 1

t is in Quadrant III.

Explanation:

First, find cos t by using trig identity:
`cos^2 x = 1/(1 +tan^2 x)` -->
`cos^2 t = 1/(1 + 25/144) = 144/169` --> `cos t = +- 12/13`
Since t is in Quadrant II, cos t is negative. Take the negative value.
Find sin t.
`sin^2 t = 1 - cos^2 t = 1 - 144/169 = 25/169` --> `sin t = +- 5/13`
Since t is in Quadrant II, then sin t is positive.
`sin 2t = 2 sin t.cos t = 2(5/13)(-12/13) = - 120/169`
`cos 2t = 1 - 2sin^2 t = 1 - 50/169 = 119/169`
`tan 2t = (sin 2t)/(cos 2t) = (- 120/169)(169/119) = - 120/119` (1)
You may check the results by calculator.
`tan t = - 5/12` -->`t = - 22^@ 62 + 180^@ = 157^@38` (Quadrant II)
`tan 2t = tan 314^@76 = - 1.01`
`2t = 314^@76` is in Quadrant III. Compare to answer (1) -->
`tan 2t = - 120/119 = - 1.01`. OK