Finding cos2*theta

**IBstudent** · June 25th 2011, 11:17 PM

Hey,
I have a problem that I couldn't solve, I have no idea of the method I shuold use to solve it, please anyone who does can you show me?

The question is attached...

Thanks

**BAdhi** · June 25th 2011, 11:36 PM

draw a rectangled triangle and mark an angle as $2\theta$ and mark appropriate lengths of the sides which will give $\sin{2\theta}=\frac{4\sqrt{5}}{9}$ and find the unknown side so that you can find $\cos{2\theta}$ . this is same as using the identity $\cos^2{\theta}=1-\sin^2{\theta}$

**IBstudent** · June 25th 2011, 11:39 PM

Originally Posted by BAdhi

draw a rectangled triangle and mark an angle as $2\theta$ and mark appropriate lengths of the sides which will give $\sin{2\theta}=\frac{4\sqrt{5}}{9}$ and find the unknown side so that you can find $\cos{2\theta}$ . this is same as using the identity $\cos^2{\theta}=1-\sin^2{\theta}$

I'm sorry, but I don't understand you wrote....
Thanks for the attempt though :-)

**BAdhi** · June 26th 2011, 12:06 AM

either use the identity $\cos^2{2\theta}=1-\sin^2{2\theta}$ or

use the following triangle to find $\cos{2\theta}$

**IBstudent** · June 26th 2011, 12:32 AM

I tried your triangle method and got 2/9, is this correct?

**Joanna** · June 26th 2011, 12:48 AM

Originally Posted by IBstudent

I tried your triangle method and got 2/9, is this correct?

No this isn't correct, but it's close. The identity that BAdhi posted is one of the most useful in Trigonometry, and really should be memorized. Knowing it will make much of the computation you end up doing so much easier.

**Soroban** · June 26th 2011, 05:33 AM

Hello, IBstudent!

$\sin2\theta \:=\:\frac{4\sqrt{5}}{9}$

$\text{Find the exact value of }\cos2\theta.$

$\text{Recall: }\:\sin^2\!x + \cos^2\!x \:=\:1$

So we have: . $\sin^2\!2\theta + \cos^2\!2\theta \:=\:1$

. . . . . . . . $\left(\frac{4\sqrt{5}}{9}\right)^2 + \cos^2\!2\theta \:=\:1$

. . . . . . . . . . . . $\frac{80}{81} + \cos^2\!2\theta \:=\:1$

. . . . . . . . . . . . . . . . $\cos^2\!2\theta \:=\:\frac{1}{81}$

. . . . . . . . . . . . . . . . $\cos2\theta \:=\:\pm\frac{1}{9}$

**Prove It** · June 26th 2011, 08:03 AM

Originally Posted by IBstudent

Hey,
I have a problem that I couldn't solve, I have no idea of the method I shuold use to solve it, please anyone who does can you show me?

The question is attached...

Thanks

Are you told which quadrant of the unit circle that the angle $\displaystyle 2\theta$ lies in?

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