i think sin 200 is smaller than sin 0.. because sin 200= - sin 20.. sin 0 = 0 of course 0 > - sin 20
sin 0 = 0 cos 0 = 1
sin(2*pi) - not pie - is the same as sin(0) = 0
sin(pi) = 0
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
i think sin 200 is smaller than sin 0.. because sin 200= - sin 20.. sin 0 = 0 of course 0 > - sin 20
sin(-pi) = sin(-180) = 0 So the answer is 0
sin 0 = 0 cos 0 = 1
No. sin(0) = 0 So cos(0)*sin(0) = 0 so the left hand side = 1
sin(2*pi) - not pie - is the same as sin(0) = 0
sin(0)=0, therefore ysin(0)=0
Sine (0) = 0 Sin(30) = 0.5 Sin(45) = 0.7071... Sin(60) = 0.8660.... Sin(90) = 1 Are just a few of the Sine(Trigonometric) values.
sin(pi) = 0
lim(h→0) (sin x cos h + cos x sin h - sin x)/h As h tends to 0, both the numerator and the denominator have limit zero. Thus, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply l'Hopital's Rule and the limit equals: lim(h→0) (sin x cos h + cos x sin h - sin x)/h = lim(h→0) (sin x cos h + cos x sin h - sin x)'/h' = lim(h→0) [[(cos x)(cos h) + (sin x)(-sin h)] + [(-sin x)(sin h) + (cos x)(cos h)] - cos x]]/0 = cosx/0 = ∞
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
Sin (0) isn't a function, it's a number.