Question

How do you find the exact value of `cos2theta=2costheta-2cos^2theta` in the interval `0<=theta<2pi`?

Answer 1

`36^@, 108^@, 252^@, 324^@`

Explanation:

`cos 2t = 2cos t - 2cos^2 t`
Use trig identity: `cos 2t = 2cos^2 t - 1`
We get after replace cos 2t on the left side:
`2cos^2 t - 1 = 2cos t - 2cos^2 t`
`4cos^2 t - 2cos t - 1 = 0`
Solve this quadratic equation for cos t.
`D = d^2 = b^2 - 4ac = 4 + 16 = 20` --> `d = +- 2sqrt5`
There are 2 real roots:
`cos t = -b/(2a) +- d/(2a) = 1/4 +- sqrt5/4 = (1 +- sqrt5)/4`
a. `cos t = (1 + sqrt5)/4 = 0.8090` --> `t = +- 36^@`
b. `cos t = (1 - sqrt5)/4 = - 0.3090` --> `t = +- 108^@`
`- 36^@` is co-terminal to `324^@`
`- 108^@` is co-terminal to `252^@`
Answers for (0, 360):
`36^@, 108^@, 252^@, 324^@`