Angle A is in standard position with its terminal arm in quadrant III, the Trigonometry ratio sinA=-3/7 is given, how do you find the values of the other ratios?

Apr 29, 2015

There are, broadly speaking, two approaches.

No pictures or thinking about pictures (or not much)

${\sin}^{2} A + {\cos}^{2} A = 1$.

We are given: $\sin A = - \frac{3}{7}$, so we can calculate:

${\left(- \frac{3}{7}\right)}^{2} + {\cos}^{2} A = 1$

${\cos}^{2} A = \frac{40}{49}$

This tells us that:

$\cos A = \pm \sqrt{\frac{40}{49}} = \pm \frac{2 \sqrt{10}}{7}$

Now the fact that A has terminal side in Q III tells us the $\cos A$ is negative, so

$\cos A = - \frac{2 \sqrt{10}}{7}$

Now that we have $\sin A$ and $\cos A$, we can get the other values using:

$\tan A = \sin \frac{A}{\cos} A$, $\sec A = \frac{1}{\cos} A$ and so on.

Method 2, thinking about the definitions of the trigonometric ratios:

There are several possible approaches to this:

2a:
Think of $A$ corresponding to a point $\left(x , y\right)$ on the unit circle:
${x}^{2} + {y}^{2} = 1$

with $\sin A = y$, we see immediately that $y = - \frac{3}{7}$

Now use ${x}^{2} + {y}^{2} = 1$ to get $x = \pm \sqrt{\frac{40}{49}} = \pm \frac{2 \sqrt{10}}{7}$

And the fact that we are in Q III, to get $x = - \frac{2 \sqrt{10}}{7}$

$\cos A = x$, $\tan A = \frac{y}{x}$ and so on.

2b

for points on the terminal side $\left(x , y\right)$ with $r = \sqrt{{x}^{2} + {y}^{2}}$

and $\sin A = \frac{y}{r} = \frac{- 3}{7}$

We can choose one of these points (the unit circle approaches chooses $r = 1$)

Let's make $y = - 3$, so we must have $r = 7$ and, using $r = \sqrt{{x}^{2} + {y}^{2}}$, we get $x = \pm 2 \sqrt{10}$

In Q III, we must have $x = - 2 \sqrt{10}$

Now:

$\cos A = \frac{x}{r}$, $\tan A = \frac{y}{r}$, and so on.