Question

If sinθ = 2/3 and θ is located in Quadrant II, then tan2θ = _____.

Answer 1

Answer:

$-4 \sqrt{5}$

Step-by-step explanation:

Quadrant 2 means cosine is negative.

So $\sin(\theta)=\frac{2}{3} =\frac{\text{ opp }}{\text{ hyp }}$

So the adjacent side is $\sqrt{3^2-2^2}=\sqrt{9-4}=\sqrt{5}$

So $\cos(\theta)=-\frac{\sqrt{5}}{3}$

Now to find $\tan(2 \theta)$

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

We will need $\tan(\theta)$ before proceeding.

$\tan(\theta) =\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{2}{3}}{\frac{-\sqrt{5}}{3}}=\frac{-2}{\sqrt{5} }$

Now plug it in and the rest is algebra.

$\tan(2 \theta) =\frac{2\tan(\theta)}{1-\tan^2(\theta)} =\frac{2 (\frac{-2}{\sqrt{5}}}{1-\frac{4}{5}}$

Now the algebra, the simplifying.... We need to get rid of the compound fraction.  We will multiply top and bottom by $5 \sqrt{5}$

This will give us

$\frac{-4(5)}{5 \sqrt{5}-4 \sqrt{5}}$

$\frac{-20}{\sqrt{5}}$

Multiply top and bottom by $\sqrt{5}$

$\frac{-20 \sqrt{5}}{5}$

The answer reduces to

$-4 \sqrt{5}$