# Math Help - Solving for sin(x)

1. ## Solving for sin(x)

Given that $\tan(x) = \frac{2}{3}\mbox{ and } \pi

I know that $\pi puts $x$ in Quadrant III in the unit circle, but I don't see how that helps me, other than the fact that $\sin(x)$ will be negative. I solved for $\sin(x)$, but I'm pretty sure it isn't the right answer.

$\tan(x) = \frac{2}{3}\\\\\frac{\sin(x)}{cos(x)} = \frac{2}{3}\\\\\sin(x) = \frac{2cos(x)}{3}\\$

also, it gave a,b,c,d,e choices.

a) $\frac{\sqrt{10}}{10}$

b) $\frac{-\sqrt{13}}{2}$

c) $\frac{-\sqrt{10}}{10}$

d) $\frac{-2\sqrt{13}}{13}$

e) $\frac{2\sqrt{13}}{13}$

2. ## Re: Solving for sin(x)

If you know:
$\tan(x)=\frac{2}{3} \Leftrightarrow x=\arctan\left(\frac{2}{3}\right)$.
and you know $\tan(x)=\tan(\pi+x)$

Can you now complete? ...

3. ## Re: Solving for sin(x)

Originally Posted by Fenixx09
Given that $\tan(x) = \frac{2}{3}\mbox{ and } \pi
I know that $\pi puts $x$ in Quadrant III in the unit circle.
Recall that $\tan(t)=\frac{y}{x}\text{ and }x^2+y^2=r^2$.
Also $\sin(t)=\frac{y}{r}$.

Don't forget that in III, $y<0~\&~r>0$.

4. ## Re: Solving for sin(x)

Originally Posted by Fenixx09
Given that $\tan(x) = \frac{2}{3}\mbox{ and } \pi
reference triangle in quad III ... note the tangent value of the reference angle is $\frac{-2}{-3} = \frac{2}{3}$

calculate the value of the hypotenuse and use your basic definition of sine for the reference angle.

5. ## Re: Solving for sin(x)

Thanks guys, I really wanted to know how to do this without a calculator! And I couldn't get off of the unit circle until skeeter posted that graph :/

$(-3)^{2}+(-2)^{2}=(\mbox{hypotenuse})^{2}\\\\9+4=(\mbox{hypot enuse})^{2}\\\\13=(\mbox{hypotenuse})^{2}\\\\\sqrt {13}=(\mbox{hypotenuse})\\\\sin(x)=\frac{\mbox{(\o pposite)}}{(\mbox{hypotenuse})}\\\\sin(x)=\frac{-2}{\sqrt{13}}\\\\sin(x)=\frac{-2\sqrt{13}}{13}$

6. ## Re: Solving for sin(x)

My solution would be:
$\tan(x)=\frac{2}{3} \Leftrightarrow x=\arctan\left(\frac{2}{3}\right)$ and $x+\pi=\arctan\left(\frac{2}{3}\right) \Leftrightarrow x=\arctan\left(\frac{2}{3}\right) - \pi$

Now you just have to calculate:
$\sin\left[\arctan\left(\frac{2}{3}\right)-\pi \right]=\sin \left[\arctan\left(\frac{2}{3}\right)\right]\cdot \cos(\pi)-\sin(\pi)\cdot \cos\left[\arctan\left(\frac{2}{3}\right)\right]$

$=-\frac{\frac{2}{3}}{\sqrt{1+\left(\frac{2}{3}\right )^2}}$
$=-\frac{\frac{2}{3}}{\sqrt{\frac{13}{9}}}$
$=-\frac{2}{\sqrt{13}}=-\frac{2\sqrt{13}}{13}$

7. ## Re: Solving for sin(x)

Originally Posted by Siron
My solution would be:
$\tan(x)=\frac{2}{3} \Leftrightarrow x=\arctan\left(\frac{2}{3}\right)$ and $x+\pi=\arctan\left(\frac{2}{3}\right) \Leftrightarrow x=\arctan\left(\frac{2}{3}\right) - \pi$
Now you just have to calculate:
$\sin\left[\arctan\left(\frac{2}{3}\right)-\pi \right]=\sin \left[\arctan\left(\frac{2}{3}\right)\right]\cdot \cos(\pi)-\sin(\pi)\cdot \cos\left[\arctan\left(\frac{2}{3}\right)\right]$
$=-\frac{\frac{2}{3}}{\sqrt{1+\left(\frac{2}{3}\right )^2}}$
$=-\frac{\frac{2}{3}}{\sqrt{\frac{13}{9}}}$
$=-\frac{2}{\sqrt{13}}=-\frac{2\sqrt{13}}{13}$
I must say that approach seems over-the-top to me.
Here are some ideas you may consider, I realize are the product of a European system. This forum is titled pre-university trigonometry. Now if this poster is in North America he/she may not ever have seen the arctangent function. In fact he/she may well only 9th or 10th year student. But looking at his/her reply, it appears that skeeter’s reply was the one understood. That tells me that this poster is not even up to the analytic trigonometry that I posted.
My point is that pre-university trigonometry has many different meanings.

8. ## Re: Solving for sin(x)

You're absolutely right (what do you mean with an European product?) and probably my solution doens't interest Fenixx09 but maybe there are other members interested.