[Further Maths Level 2] Trig Identities Question

Hey guys, can you give me a helping hand with this question?

Angle θ is obtuse
Sinθ = (sqrt5)/3
Work out the values of Cosθ

The part where θ is obtuse is throwing me off a bit (sorry if it's hard to understand the question I don't know how to use latex)

Hey guys, can you give me a helping hand with this question?

Angle θ is obtuse
Sinθ = (sqrt5)/3
Work out the values of Cosθ

The part where θ is obtuse is throwing me off a bit (sorry if it's hard to understand the question I don't know how to use latex)

This is always a hard question for Level 2 FM students. You can use a few methods but you might like to use the identity:

$\sin^2 \theta + \cos^2 \theta = 1$

If you plug in the value for $\sin \theta$ and rearrange then square root, you'll be left with two possible values, one positive and one negative (remember there are positive and negative square roots).

So then how do you know which one to choose? You are told that $\theta$ is obtuse so this tells you whether $\cos \theta$ is always negative or always positive. Try drawing the $\cos$ graph and see if you can work out which one it is. Alternatively, let me know if you've heard of the CAST diagram (most Level 2 FM students haven't) and I'll explain it using that.

So looking at the Cosine graph, between 90 and 180 (obtuse) it is negative, but I must've gone wrong somewhere because one of my answers is positive and the other is less than -1 (which can't be on the graph?)

I did:
1 - (sqrt5)/3
and
-1 - (sqrt5/3)

are equal to Cos^2θ

You've made a mistake in your working. The identity you need is

$\cos^2 \theta = 1 - \sin^2 \theta$

You have $\sin \theta = \frac{\sqrt{5}}{3} \ \$ so $\ \ \sin^2 \theta = \frac{5}{9}$.

The place you need to think about positive/negative answers is when you go from $\cos^2 \theta$ to $\cos \theta$ in your working.

Have another go and post your working if you get stuck.

Excuse the handwriting is this right?

Perfect