Quick Question - Trigonometry? (1 Viewer)

[Zokunu]

I don't get it...

For pi/2 < x < pi, use a diagram of a unit circle to show that:

(a) cos (pi + x ) = -cosx

(b) Sin (2pi - x) = -sinx

I looked at the Fitzpatrick Solution booklet and still don't get this. Ex.15.3 Q11

Please help me out? Thanks

 

[iStudent]

Not sure if I'm right, but here's what I think:
pi/2 < x < pi is in the 2nd quadrant, therefore sine is positive and all else negative

(a) cos (pi + x ) = -cosx

rotating 180 degrees from the 2nd quadrant, we land in the 4th quadrant - cos is positive
cos has moved from a negative to a positive region
therefore, cos(pi+x) = -cosx (as their signs are different)

(b) Sin (2pi - x) = -sinx

flipping on the x axis then revolving 360 degrees, we land in the 3rd quadrant
in this sine is negative
therefore sine had its sign changed from positive to negative
so sin(2pi-x) = -sinx

 

[Zokunu]

Not sure if I'm right, but here's what I think:
pi/2 < x < pi is in the 2nd quadrant, therefore sine is positive and all else negative

(a) cos (pi + x ) = -cosx

rotating 180 degrees from the 2nd quadrant, we land in the 4th quadrant - cos is positive
cos has moved from a negative to a positive region
therefore, cos(pi+x) = -cosx (as their signs are different)

(b) Sin (2pi - x) = -sinx

flipping on the x axis then revolving 360 degrees, we land in the 3rd quadrant
in this sine is negative
therefore sine had its sign changed from positive to negative
so sin(2pi-x) = -sinx

Thanks man