Odd quadrant information?

Dec 2011
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0
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
Feb 2009
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West Midlands, England
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
Aug 2006
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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 \(\displaystyle 0<\theta<\pi\)