1. ## Trig

If $cos\beta = -\frac{3}{11}$ and $\beta$ is in the second quadrant, find the exact value of $sin\;2\beta$

Well $sin2\beta = 2sin\beta cos\beta$

My final answer is $-5.77$ is this correct?

2. Originally Posted by R3ap3r
If $cos\beta = -\frac{3}{11}$ and $\beta$ is in the second quadrant, find the exact value of $sin\;2\beta$

Well $sin2\beta = 2sin\beta cos\beta$

My final answer is $-5.77$ is this correct?
Hullo,

A sine can barely be <-1 :/

Here is how to do. You know that $\beta$ is in the second quadrant, so $\sin(\theta)>0$ and $\cos(\theta)<0$

We know that $\cos^2(\beta)+\sin^2(\beta)=1$

--> $\sin^2(\beta)=1-\left(-\frac{3}{11}\right)^2=\frac{112}{121}$

Since $\sin(\beta)>0$, $\sin(\beta)=\sqrt{\frac{112}{121}} \approx \dots$

Hence, $\sin(2 \beta)=\dots$

3. Well I know atleast what i did wrong. i forgot to put $\sqrt{112}$ over 11 in the formula.

New answer is $-.525$ hows this look?

4. Originally Posted by janvdl
Code:
   |\
| \
|  \  sqrt{130}
11 |   \
|__o_\

3
That $o$ there represents Beta.

We wish to find $2 Sin \beta Cos \beta$

= $2 \left( \frac{11}{ \sqrt{130} } \right) \left( \frac{3}{ \sqrt{130} } \right)$

= $\frac{33}{65}$
I believe you made an error. For one its the $\sqrt{112}$ and my math is $2\cdot \left(\frac{\sqrt{112}}{11}\right)\left(\frac{-3}{11}\right)$

whose right? lol

5. Originally Posted by R3ap3r
I believe you made an error. For one its the $\sqrt{112}$ and my math is $2\cdot \left(\frac{\sqrt{112}}{11}\right)\left(\frac{-3}{11}\right)$

whose right? lol
I'm pretty sure it's this one

6. In Janvdl's diagram, the 11 should be the hypotenuse.

7. Originally Posted by R3ap3r
I believe you made an error. For one its the $\sqrt{112}$ and my math is $2\cdot \left(\frac{\sqrt{112}}{11}\right)\left(\frac{-3}{11}\right)$

whose right? lol
Moo's. And I'm going to bed because I can't seem to concentrate anymore. Sorry for the mistake Reap3r.

8. np