I'm fairly new to more complex trigonometry, and i've found myself stumped on a certain question which can be found in exercise 6C of the Cambridgbe Advanced Mathematics course textbook. The question reads as follows...
If Sin A = 2/3, and A is obtuse, find the exact values of Cos A, Sin 2A and tan 2A.
I know how to use the double angle formula to calculate Sin 2A and tan 2A, but i'm stuck on trying to find Cos A which i need to put into the double angle formula to find the last two parts...
Any ideas? Thanks
Would i start off using the Pythagorean identity:
(Sin A)^2 + (Cos A)^2 = 1
let Sin A = 2/3
(2/3)^2 + (Cos A)^2 = 1
take (2/3)^2 from each side...
(Cos A)^2 = 1 - 4/9
because Sin A is Obtuse, the square root of the RHS must be negative (Deduced from Sin / Cos graphs)
Cos A = - [SQRT]5 / 3
so does that work?