How to find sinθ given sin2θ?

**michael710** · July 13th 2010, 07:58 PM

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.

**Prove It** · July 13th 2010, 08:15 PM

First of all, if $\frac{3\pi}{2} < 2\theta < 2\pi$ , then $\frac{3\pi}{4} < \theta < \pi$ .

Since this is in Quadrant 2, you know $0 < \sin{\theta} < 1$ .

You have $\sin{2\theta} = -\frac{2}{3}$

$2\sin{\theta}\cos{\theta} = - \frac{2}{3}$

$2\sin{\theta}\sqrt{1 - \sin^2{\theta}} = -\frac{2}{3}$

$(2\sin{\theta}\sqrt{1 - \sin^2{\theta}})^2 = \left(-\frac{2}{3}\right)^2$

$4\sin^2{\theta}(1 - \sin^2{\theta}) = \frac{4}{9}$

$\sin^2{\theta}(1 - \sin^2{\theta}) = \frac{1}{9}$

$\sin^2{\theta} - \sin^4{\theta} = \frac{1}{9}$

$\sin^4{\theta} - \sin^2{\theta} + \frac{1}{9} = 0$

$X^2 - X + \frac{1}{9} = 0$ where $X = \sin^2{\theta}$

$X^2 - X + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + \frac{1}{9} = 0$

$\left(X - \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{1}{9} = 0$

$\left(X - \frac{1}{2}\right)^2 - \frac{5}{36} = 0$

$\left(X - \frac{1}{2}\right)^2 = \frac{5}{36}$

$X - \frac{1}{2} = \frac{\pm \sqrt{5}}{6}$

$X = \frac{1}{2} \pm \frac{\sqrt{5}}{6}$

$X = \frac{3 \pm \sqrt{5}}{6}$

$\sin^2{\theta} = \frac{3 \pm \sqrt{5}}{6}$

$\sin{\theta} = \frac{\sqrt{3 \pm \sqrt{5}}}{\sqrt{6}}$

$\sin{\theta} = \frac{\sqrt{6}\sqrt{3 \pm \sqrt{5}}}{6}$

$\sin{\theta} = \frac{\sqrt{6(3 \pm \sqrt{5})}}{6}$

$\sin{\theta} = \frac{\sqrt{18 \pm 6\sqrt{5}}}{6}$ .

The only solution that lies in $0 < \sin{\theta} < 1$ is

$\sin{\theta} = \frac{\sqrt{18 - 6\sqrt{5}}}{6}$ .

**CaptainBlack** · July 13th 2010, 08:22 PM

Originally Posted by Prove It

First of all, if $\frac{3\pi}{2} < 2\theta < 2\pi$ , then $\frac{3\pi}{4} < \theta < \pi$ .

Since this is in Quadrant 2, you know $0 < \sin{\theta} < 1$ .

You have $\sin{2\theta} = -\frac{2}{3}$

$2\sin{\theta}\cos{\theta} = - \frac{2}{3}$

$2\sin{\theta}\sqrt{1 - \sin^2{\theta}} = -\frac{2}{3}$

What sign is $\cos(\theta)$ in quadrant 2?

CB

**michael710** · July 13th 2010, 08:29 PM

Ouch. I need to work on my algebra. It wouldn't have occurred to me to do it that way at all.

It makes sense I think, but in:

, is there a reason (-1/2) was used, as opposed to any other number?

**Prove It** · July 13th 2010, 08:30 PM

Originally Posted by CaptainBlack

What sign is $\cos(\theta)$ in quadrant 2?

CB

$^{-}ve$ .

But $\sin{\theta}$ is $^{+}ve$ , so $2\sin{\theta}\cos{\theta}$ is $^{-}ve$ .

I suppose I should have written $-2\sin{\theta}\sqrt{1 - \sin^2{\theta}}$ , but it doesn't really matter once you square both sides.

**Prove It** · July 13th 2010, 08:32 PM

Originally Posted by michael710

Ouch. I need to work on my algebra. It wouldn't have occurred to me to do it that way at all.

It makes sense I think, but in:

, is there a reason (-1/2) was used, as opposed to any other number?

Yes, when you complete the square in a quadratic, you add and subtract half of the square of the $b$ value. In this case the $b$ value is $-1$ .

**michael710** · July 13th 2010, 08:55 PM

I'm not even sure I've ever covered completing the square, but the concept makes sense, and that's the only place I questioned. I'll drill this into my brain until I can do it myself.

Thanks! And thanks for such a quick response.

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