trig
posted by Cody on .
How do I find the exact value of sin2x when secx is equal to the negative square root of three

then cosx = 1/√3
x must be in II or III
in II:
sinx = √2/√3
sin 2x = 2sinxcosx = 2(√2/√3)(1/√3) = 2√2/3
in III:
sinx = √2/√3
sin 2x = +2√2/3