problem in addition formulae and double angle formulae

• July 5th 2010, 05:15 AM
mastermin346
problem in addition formulae and double angle formulae

given $tan2\theta=\frac{4}{3},$ without using calculator,find the values of $sin\theta$.
• July 5th 2010, 05:43 AM
eumyang
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 $sin 2\theta=\frac{4}{3},$ without using calculator,find the values of $sin\theta$.
• July 5th 2010, 05:48 AM
mastermin346
yes..that is the question.

opp..sory..the question is tan 2\theta..i have edit it
• July 5th 2010, 06:17 AM
eumyang
$sin\,\theta= \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}$
I started by using the tangent double identity:
$\frac{2tan\,\theta}{1-tan^2\,\theta}=\frac{4}{3}$
Cross-multiplied:
$3\left(2tan\,\theta \right)=4\left(1-tan^2\,\theta \right)$
... solved for $tan\,\theta$, then rewrote in terms of $sin\,\theta$.

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

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

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

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

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

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

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

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

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

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

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

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

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

Case 1:

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

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

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

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

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

Apply a similar process for Case 2.