please help me with this question:

given $\displaystyle tan2\theta=\frac{4}{3},$ without using calculator,find the values of $\displaystyle sin\theta$.

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- Jul 5th 2010, 05:15 AMmastermin346problem in addition formulae and double angle formulae
please help me with this question:

given $\displaystyle tan2\theta=\frac{4}{3},$ without using calculator,find the values of $\displaystyle sin\theta$. - Jul 5th 2010, 05:43 AMeumyang
Uh... come again? Maybe I'm not seeing it, but is there more to this problem? It this part of a bigger problem?

EDIT: Do you mean this:

Quote:

given $\displaystyle sin 2\theta=\frac{4}{3},$ without using calculator,find the values of $\displaystyle sin\theta$.

- Jul 5th 2010, 05:48 AMmastermin346
yes..that is the question.

opp..sory..the question is tan 2\theta..i have edit it - Jul 5th 2010, 06:17 AMeumyang
I'm getting

**2**answers:

$\displaystyle sin\,\theta= \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}$

I started by using the tangent double identity:

$\displaystyle \frac{2tan\,\theta}{1-tan^2\,\theta}=\frac{4}{3}$

Cross-multiplied:

$\displaystyle 3\left(2tan\,\theta \right)=4\left(1-tan^2\,\theta \right)$

... solved for $\displaystyle tan\,\theta$, then rewrote in terms of $\displaystyle sin\,\theta$.

EDIT: I originally thought that both answers I wrote above were plus-minus, but I forgot that $\displaystyle tan\,2\theta$ is positive, which means that the angle $\displaystyle 2\theta$ has to be in Quadrants I or III. So $\displaystyle \theta$ is between 0 and 45° or between 90° and 135°, and for those values of $\displaystyle \theta$, sine is positive. - Jul 5th 2010, 06:20 AMProve It
You should know that

$\displaystyle \tan{2\theta} = \frac{2\tan{\theta}}{1 - \tan^2{\theta}}$.

So if $\displaystyle \tan{2\theta} = \frac{4}{3}$ then

$\displaystyle \frac{2\tan{\theta}}{1 - \tan^2{\theta}} = \frac{4}{3}$

$\displaystyle 3(2\tan{\theta}) = 4(1 - \tan^2{\theta})$

$\displaystyle 6\tan{\theta} = 4 - 4\tan^2{\theta}$

$\displaystyle 4\tan^2{\theta} + 6\tan{\theta} - 4 = 0$

$\displaystyle 2\tan^2{\theta} + 3\tan{\theta} - 2 = 0$

$\displaystyle 2\tan^2{\theta} - \tan{\theta} + 4\tan{\theta} - 2 = 0$

$\displaystyle \tan{\theta}(2\tan{\theta} - 1) + 2(2\tan{\theta} - 1) = 0$

$\displaystyle (2\tan{\theta} - 1)(\tan{\theta} + 2) = 0$

$\displaystyle 2\tan{\theta} - 1 = 0$ or $\displaystyle \tan{\theta} + 2 = 0$

$\displaystyle \tan{\theta} = \frac{1}{2}$ or $\displaystyle \tan{\theta} = -2$.

Case 1:

$\displaystyle \tan{\theta} = \frac{1}{2}$.

This means you are dealing with a right-angle triangle of side lengths $\displaystyle O = 1$ and $\displaystyle A = 2$.

By Pythagoras, that means that $\displaystyle H = \sqrt{5}$.

Since $\displaystyle \sin{\theta} = \frac{O}{H}$ that means

$\displaystyle \sin{\theta} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$.

Apply a similar process for Case 2.