Question

How do you find the exact values of Tan (-195) using the half angle formula?

Answer 1

Fin tan (-195)

Ans: `sqrt3 - 2`

Explanation:

Call tan (-195) = t
`tan (- 390) = 2t = tan (-30 - 360) = tan (-30) = - tan (30) = - 1/sqrt3`
Apply trig identity: `tan 2t = (2t)/(1 - t^2)`
`-1/sqrt3 = (2t)/(1 - t^2)`
`-1 + t^2 = 2sqrt3t.` Solve the quadratic equation:
`t^2 - 2sqrt3t - 1 = 0`
`D = d^2 = b^2 - 4ac = 12 + 4 = 16` --> `d = +-4`
The 2 real roots are:
`t = sqrt3 +- 2`.
Since the arc (-195) is located in Quadrant II, its tan is positive, then:
`tan (-195) = t = sqrt3 - 2`

Check by calculator:
tan (-195) = -0.27
`sqrt3 - 2` = 1.73 - 2 = - 0.27. OK