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 :)

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

Quote:

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?

$\displaystyle \cos^2{A} + \sin^2{A} = 1$

solving for $\displaystyle \cos{A}$ ...

$\displaystyle \cos{A} = \pm \sqrt{1 - \sin^2{A}}$

since A is obtuse, the value of $\displaystyle \cos{A}$ would be negative (why?)

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

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? :)

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