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?

Question

6997 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-29T19:50:31

There are, broadly speaking, two approaches.

No pictures or thinking about pictures (or not much)

$sin^2A + cos^2 A = 1$ .

We are given: $sinA = -3/7$ , so we can calculate:

$(-3/7)^2 + cos^2A = 1$

$cos^2A = 40/49$

This tells us that:

$cosA = +- sqrt(40/49) = +- (2sqrt10)/7$

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

$cosA = - (2sqrt10)/7$

Now that we have $sinA$ and $cosA$ , we can get the other values using:

$tanA = sinA/cosA$ , $secA = 1/cosA$ 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 $(x, y)$ on the unit circle:
$x^2+y^2=1$

with $sinA = y$ , we see immediately that $y=-3/7$

Now use $x^2+y^2=1$ to get $x= +- sqrt(40/49) = +- (2sqrt10)/7$

And the fact that we are in Q III, to get $x= - (2sqrt10)/7$

$cosA = x$ , $tanA = y/x$ and so on.

2b

for points on the terminal side $(x,y)$ with $r = sqrt(x^2+y^2)$

and $sinA = y/r = (-3)/7$

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 $x=+- 2sqrt10$

In Q III, we must have $x=- 2sqrt10$

Now:

$cosA = x/r$ , $tanA=y/r$ , and so on.