Thread: Double Angle Formula (Sin 2A, Cos 2A...)

Hi,

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

Originally Posted by lrwatton93

Hi,

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

familiar with this identity?



solving for ...



since A is obtuse, the value of would be negative (why?)

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?

that's all, folks.