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
draw a rectangled triangle and mark an angle as $\displaystyle 2\theta$ and mark appropriate lengths of the sides which will give $\displaystyle \sin{2\theta}=\frac{4\sqrt{5}}{9}$ and find the unknown side so that you can find $\displaystyle \cos{2\theta}$. this is same as using the identity $\displaystyle \cos^2{\theta}=1-\sin^2{\theta}$
Hello, IBstudent!
$\displaystyle \sin2\theta \:=\:\frac{4\sqrt{5}}{9}$
$\displaystyle \text{Find the exact value of }\cos2\theta.$
$\displaystyle \text{Recall: }\:\sin^2\!x + \cos^2\!x \:=\:1$
So we have: .$\displaystyle \sin^2\!2\theta + \cos^2\!2\theta \:=\:1$
. . . . . . . . $\displaystyle \left(\frac{4\sqrt{5}}{9}\right)^2 + \cos^2\!2\theta \:=\:1$
. . . . . . . . . . . . $\displaystyle \frac{80}{81} + \cos^2\!2\theta \:=\:1$
. . . . . . . . . . . . . . . .$\displaystyle \cos^2\!2\theta \:=\:\frac{1}{81}$
. . . . . . . . . . . . . . . . $\displaystyle \cos2\theta \:=\:\pm\frac{1}{9}$