Posted by **Jon** on Friday, February 15, 2008 at 4:54pm.

Find the exact value of sin2(theta) if cos(theta) = -sqrt 5/3 and 180 < theta < 270.

A)-1/9

B)-4 sqrt 5/9

C)1/9

D)4 sqrt 5/9

B?

sin^2(theta) + cos^2(theta) = 1

sin^2(theta) = 1 - cos^2(theta)

sin^2(theta) = 1 - (sqrt 5/3)^2

sin^2(theta) = 1 - (sqrt 25/9)

- Trig -
**Reiny**, Friday, February 15, 2008 at 5:07pm
You are getting a lot of these wrong lately.

(I am going to use sinx for your sin(theta)

since cosx = √5 /3

using Pythagoras I found the other side ot the triangle to be 2.

But we are in the third quadrant so sinx =-2/3

then sin 2x = 2sinxcosx

= 2(-2/3)(-√5/3)

= 4√5/3 which is D

with a calculator, you could have easily checked that your choice and the others beside D would not work.

- Trig -
**Jon**, Friday, February 15, 2008 at 5:15pm
I really thought I had that one.

My 1st thought was D but I figured since sqrt 5/3 was negative then so would my final answer.

- Trig -
**Jon**, Friday, February 15, 2008 at 5:26pm
No

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