Question

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?

Answer 1

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.