# problem in addition formulae and double angle formulae

#### mastermin346

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

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

#### mastermin346

yes..that is the question.

opp..sory..the question is tan 2\theta..i have edit it

#### eumyang

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

• mastermin346

#### Prove It

MHF Helper
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.

• mastermin346