Clear working of $\sin(\arccos(-\frac13))$

Question

Can someone show this please. I always had trouble with this in A Level.

It is strange that this expression as an exact value because there is no exact value of $\arccos(-\frac13)$.

Draw a triangle and mark on the "arccos($-\frac{1}{3}$)" angle, then what would the sin of it be — Alec Teal, Mar 08 '15 at 13:12

score 4 · Accepted Answer · answered Mar 08 '15 at 13:11

4

Hint: put $\arccos (-{1\over 3})=\alpha$. Then $\cos\alpha =-{1\over 3}$. Can you calculate $\sin \alpha$?

answered Mar 08 '15 at 13:11

Fermat

1

oh yeah, just square both sides and use identities. thanks – snowman Mar 08 '15 at 13:18
Well done! @snowman – Fermat Mar 08 '15 at 13:20
you shouldn't congratulate me since I am at uni and I should have known this from the start. – snowman Mar 08 '15 at 13:22

1 Answers1