#### 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?

#### e^(i*pi)

MHF Hall of Honor
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

#### Plato

MHF Helper
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 $$\displaystyle 0<\theta<\pi$$