please help me with this question:
given $\displaystyle tan2\theta=\frac{4}{3},$ without using calculator,find the values of $\displaystyle sin\theta$.
please help me with this question:
given $\displaystyle tan2\theta=\frac{4}{3},$ without using calculator,find the values of $\displaystyle sin\theta$.
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:
given $\displaystyle sin 2\theta=\frac{4}{3},$ without using calculator,find the values of $\displaystyle sin\theta$.
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.
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.