problem in addition formulae and double angle formulae

Oct 2009
202
4
please help me with this question:

given \(\displaystyle tan2\theta=\frac{4}{3},\) without using calculator,find the values of \(\displaystyle sin\theta\).
 
Last edited:
Jan 2010
278
138
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 \(\displaystyle sin\theta\).
 
Oct 2009
202
4
yes..that is the question.

opp..sory..the question is tan 2\theta..i have edit it
 
Jan 2010
278
138
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.
 
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Prove It

MHF Helper
Aug 2008
12,897
5,001
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.
 
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