Question

The angle 0 lies in Quadrant II . Cos 0 = ormula1" title=" - \frac{2}{3} " alt=" - \frac{2}{3} " align="absmiddle" class="latex-formula">
What is tan 0 ?

Answer 1

Answer:

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

Step-by-step explanation:

Since we are in quadrant 2, sine is positive.  Since sine is positive and cosine is negative, then tangent is negative.

Now I'm going to find the sine value of this angle given using one of the Pythagorean Identities, namely $\sin^2(\theta)+\cos^2(\theta)=1$.

If given $\cos(\theta)=\frac{-2}{3}$, then we have $\sin^2(\theta)+(\frac{-2}{3})^2=1$ by substitution of $\cos(\theta)=\frac{-2}{3}$.

Let's solve:

$\sin^2(\theta)+(\frac{-2}{3})^2=1$ for $\sin(\theta)$.

$\sin^2(\theta)+\frac{4}{9}=1$

Subtract 4/9 on both sides:

$\sin^2(\theta)=1-\frac{4}{9}$

Simplify:

$\sin^2(\theta)=\frac{5}{9}$

Square root both sides:

$\sin(\theta)=\sqrt{\frac{5}{9}}$

$\sin(\theta)=\frac{\sqrt{5}}{\sqrt{9}}$

$\sin(\theta)=\frac{\sqrt{5}}{3}$

===========

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

Multiplying top and bottom by 3 gives:

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

I'm going to move the factor of -1 to the top:

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