I have spent hours trying to figure out this problem. I know how to get sin2θ, given sinθ, but not the other way around:
Find sinθ, given that sin2θ = -2/3, (3pi/2) < 2θ < 2pi.
The answer is sq(18 - 6sq(5)) / 6.
Thanks for any help.
I have spent hours trying to figure out this problem. I know how to get sin2θ, given sinθ, but not the other way around:
Find sinθ, given that sin2θ = -2/3, (3pi/2) < 2θ < 2pi.
The answer is sq(18 - 6sq(5)) / 6.
Thanks for any help.
First of all, if $\displaystyle \frac{3\pi}{2} < 2\theta < 2\pi$, then $\displaystyle \frac{3\pi}{4} < \theta < \pi$.
Since this is in Quadrant 2, you know $\displaystyle 0 < \sin{\theta} < 1$.
You have $\displaystyle \sin{2\theta} = -\frac{2}{3}$
$\displaystyle 2\sin{\theta}\cos{\theta} = - \frac{2}{3}$
$\displaystyle 2\sin{\theta}\sqrt{1 - \sin^2{\theta}} = -\frac{2}{3}$
$\displaystyle (2\sin{\theta}\sqrt{1 - \sin^2{\theta}})^2 = \left(-\frac{2}{3}\right)^2$
$\displaystyle 4\sin^2{\theta}(1 - \sin^2{\theta}) = \frac{4}{9}$
$\displaystyle \sin^2{\theta}(1 - \sin^2{\theta}) = \frac{1}{9}$
$\displaystyle \sin^2{\theta} - \sin^4{\theta} = \frac{1}{9}$
$\displaystyle \sin^4{\theta} - \sin^2{\theta} + \frac{1}{9} = 0$
$\displaystyle X^2 - X + \frac{1}{9} = 0$ where $\displaystyle X = \sin^2{\theta}$
$\displaystyle X^2 - X + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + \frac{1}{9} = 0$
$\displaystyle \left(X - \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{1}{9} = 0$
$\displaystyle \left(X - \frac{1}{2}\right)^2 - \frac{5}{36} = 0$
$\displaystyle \left(X - \frac{1}{2}\right)^2 = \frac{5}{36}$
$\displaystyle X - \frac{1}{2} = \frac{\pm \sqrt{5}}{6}$
$\displaystyle X = \frac{1}{2} \pm \frac{\sqrt{5}}{6}$
$\displaystyle X = \frac{3 \pm \sqrt{5}}{6}$
$\displaystyle \sin^2{\theta} = \frac{3 \pm \sqrt{5}}{6}$
$\displaystyle \sin{\theta} = \frac{\sqrt{3 \pm \sqrt{5}}}{\sqrt{6}}$
$\displaystyle \sin{\theta} = \frac{\sqrt{6}\sqrt{3 \pm \sqrt{5}}}{6}$
$\displaystyle \sin{\theta} = \frac{\sqrt{6(3 \pm \sqrt{5})}}{6}$
$\displaystyle \sin{\theta} = \frac{\sqrt{18 \pm 6\sqrt{5}}}{6}$.
The only solution that lies in $\displaystyle 0 < \sin{\theta} < 1$ is
$\displaystyle \sin{\theta} = \frac{\sqrt{18 - 6\sqrt{5}}}{6}$.
$\displaystyle ^{-}ve$.
But $\displaystyle \sin{\theta}$ is $\displaystyle ^{+}ve$, so $\displaystyle 2\sin{\theta}\cos{\theta}$ is $\displaystyle ^{-}ve$.
I suppose I should have written $\displaystyle -2\sin{\theta}\sqrt{1 - \sin^2{\theta}}$, but it doesn't really matter once you square both sides.