Cos(120) = -0.5
A Quadrantal angle is an angle that is not in Quadrant I. Consider angle 120. You want to find cos(120) . 120 lies in quadrant II. Also, 120=180-60. So, it is enough to find cos(60) and put the proper sign. cos(60)=1/2. Cosine is negative in quadrant II, Therefore, cos(120) = -1/2.
negative one half
cos(30)cos(55)+sin(30)sin(55)=cos(30-55) = cos(-25)=cos(25) Note: cos(a)=cos(-a) for any angle 'a'. cos(a)cos(b)+sin(a)sin(b)=cos(a-b) for any 'a' and 'b'.
Provided that any denominator is non-zero, sin = sqrt(1 - cos^2)tan = sqrt(1 - cos^2)/cos sec = 1/cos cosec = 1/sqrt(1 - cos^2) cot = cos/sqrt(1 - cos^2)
Cos(120) = -0.5
A Quadrantal angle is an angle that is not in Quadrant I. Consider angle 120. You want to find cos(120) . 120 lies in quadrant II. Also, 120=180-60. So, it is enough to find cos(60) and put the proper sign. cos(60)=1/2. Cosine is negative in quadrant II, Therefore, cos(120) = -1/2.
A Quadrantal angle is an angle that is not in Quadrant I. Consider angle 120. You want to find cos(120) . 120 lies in quadrant II. Also, 120=180-60. So, it is enough to find cos(60) and put the proper sign. cos(60)=1/2. Cosine is negative in quadrant II, Therefore, cos(120) = -1/2.
cos(180°) = -1, so it is not always positive.
negative one half
sin(-120)=sqrt(3)/2 cos(-120)=-1/2 tan(-120)=-sqrt(3) csc(-120)=2/sqrt(3) sec(-120)=-2 cot(-120)=-1/sqrt(3)
sin(t) = 7/13 cos2(t) = 1 - sin2(t) = (169 - 49)/169 = 120/169 so cos(t) = ±sqrt(120)/13. But sin(t) > 0, tan(t) < 0 implies t is in the second quadrant so cos(t) = -sqrt(120)/13 And then tan(t) = sin(t)/cos(t) = -7/sqrt(120) = -0.6390 (approx).
cos 25 = groud speed/120 ground speed = 108.76 mi/h
2cos2x - cosx -1 = 0 Factor: (2cosx + 1)(cosx - 1) = 0 cosx = {-.5, 1} x = {...0, 120, 240, 360,...} degrees
Cos times Cos
No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)
3cos