Hello,

I have a question here that reads: If angle theta lies in quadrant iii & costheta= -2/root29 then determine the exact values of the other 2 trig ratios.

When I take cos-1 of the above I get 111 degrees which would be in quadrant 2... Is that a mistake or am I doing something wrong?

Thanks! :)

nah nah. No inverse functions needed. In fact, you have to keep in mind that the principal values for arccos(x) are 0 to 180, which does not include points in QIII. So, you should have taken arccos(2/√29) = 68°. So, in QIII your angle would be 180+68=248°

Draw the angle, terminating in the point (x,y) where
x = -2
y = -5
r = √29
recall that
sinθ = y/r
cosθ = x/r
tanθ = y/x
Now it's a cinch.

Hello,

No, you're not doing anything wrong. There seems to be a mistake in the given information.

If the angle theta lies in quadrant III, the cosine of theta should be negative because cosine is negative in quadrant III.

Given that cos(theta) = -2/√29, we can determine the values of the other two trig ratios using the Pythagorean Identity: sin^2(theta) + cos^2(theta) = 1.

Here's how you can solve it step by step:

Step 1: Start with the Pythagorean Identity:
sin^2(theta) + cos^2(theta) = 1

Step 2: Substitute the given value of cos(theta):
sin^2(theta) + (-2/√29)^2 = 1

Step 3: Simplify:
sin^2(theta) + 4/29 = 1

Step 4: Rearrange the equation and get the value of sin(theta):
sin^2(theta) = 1 - 4/29
sin^2(theta) = (29 - 4)/29
sin^2(theta) = 25/29

Step 5: Take the square root of both sides to get the value of sin(theta):
sin(theta) = ±√(25/29)
sin(theta) = ±5/√29
Since theta lies in quadrant III, sin(theta) is negative, so we take the negative value:
sin(theta) = -5/√29

Step 6: Determine the value of tan(theta):
tan(theta) = sin(theta) / cos(theta)
tan(theta) = (-5/√29) / (-2/√29)
tan(theta) = 5/2

So, the exact values of the other two trig ratios are sin(theta) = -5/√29 and tan(theta) = 5/2.

I hope this helps! If you have any further questions, feel free to ask.