tanx = - sqrt(15) 90 degrees < x < 180 degrees Calculate sinx and cosx, using the calculator is forbidden. I actually know the results (sinx = sqrt(15)/4 and cosx = -1/4), but desperately need the procedure. Please help, if you can.

tanx = sinx/cosx ===> sinx = -sqrt(15)*cosx cosx^2 + sinx^2 = 1 cosx^2 + (-sqrt(15)*cosx)^2 = 1 cosx^2 + 15*cosx^2 = 1 16*cosx^2 = 1 cosx^2 = 1/16 cosx = sqrt(1/16) = 1/4 From this, you obtain sinx by "sinx^2 = 1 - cosx^2" Edit: As 90 < x < 180, cosx should be negative, so cosx = -1/4, not 1/4, sorry.

Oooh, can I try? tan(x) = -sqrt(15) square(tan(x)) = 15 square(sin(x))/square(cos(x)) = 15; square(sin(x)) = 15square(cos(x)) Since square(sin(x)) + square(cos(x)) = 1 then: square(sin(x)) = 15(1-square(sin(x))) 16square(sin(x)) = 15 square(sin(x)) = 15/16 Square(cos(x)) = (1 - 15/16) EDIT: Yeah, not done yet... oops Square(sin(x)) = 15/16 ==> sin(x) = +/- sqrt(15)/4 square(cos(x)) = 1/16 ==> cos(x) = +/- 1/4 If x is from the second quadrant (180>x>90), then sin(x) > 0, cos(x) < 0 Meaning, sin(x) = sqrt(15)/4; cos(x) = -1/4.

ugh... trigonometric identities... got a D because of these and integration redoing maths Alevel this year any tips? i tend to lose track of the whole process pretty quick

Wow, thanks guys! The solution was right before my nose and I spent three hours. How the heck didn't I figure it out by myself? Anyways, I owe you both a beer or a dosen.