# Odd quadrant information?

• Dec 11th 2011, 02:17 PM
Nosophoros57
I understand that if Cosθ is say 2/3, and 0<θ<pi/2 then both x and y are positive since that limit shows the angle is in quadrant 1.

However, in that same context, SinB is 3/5, and -pi/2<B<0 what does that mean?

why is the pi/2 now on the left and negative? What does that do to X, Y, and R?
• Dec 11th 2011, 02:45 PM
e^(i*pi)
Re: Odd quadrant information?
Quote:

Originally Posted by Nosophoros57
I understand that if Cosθ is say 2/3, and 0<θ<pi/2 then both x and y are positive since that limit shows the angle is in quadrant 1.

However, in that same context, SinB is 3/5, and -pi/2<B<0 what does that mean?

why is the pi/2 now on the left and negative? What does that do to X, Y, and R?

Consider a triangle constructed in the unit circle with angle $\displaystyle \theta$ and hypotenuse of the radius hence 1.

From basic trig rations you know that:
• $\displaystyle \sin(\theta) = \frac{opp}{adj} = \frac{y}{r} = y$
• $\displaystyle \cos(\theta) = \frac{x}{r} = x$
• $\displaystyle \tan(\theta) = \frac{y}{x}$

Therefore the quadrant will determine the sign of x and y and hence of sin,cos and tan
• Dec 11th 2011, 02:47 PM
Plato
Re: Odd quadrant information?
Quote:

Originally Posted by Nosophoros57
I understand that if Cosθ is say 2/3, and 0<θ<pi/2 then both x and y are positive since that limit shows the angle is in quadrant 1.
However, in that same context, SinB is 3/5, and -pi/2<B<0 what does that mean? why is the pi/2 now on the left and negative? What does that do to X, Y, and R?

Actually you have that all wrong.
If $\displaystyle \cos(\theta)>0$ then $\displaystyle -\tfrac{\pi}{2}<\theta<\tfrac{\pi}{2}$

If $\displaystyle \sin(\theta)>0$ then $\displaystyle 0<\theta<\pi$