0
sin(0) is 0
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
Which is smaller sin 0 or sin 200?
i think sin 200 is smaller than sin 0.. because sin 200= - sin 20.. sin 0 = 0 of course 0 > - sin 20
Sin and cosine values of 0?
sin 0 = 0 cos 0 = 1
What is the sin of 2 pie?
sin(2*pi) - not pie - is the same as sin(0) = 0
What is the sin of pi?
sin(pi) = 0
Y equals 4 sin x for x equals pi?
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
Which is smaller sin 0 or sin 200?
i think sin 200 is smaller than sin 0.. because sin 200= - sin 20.. sin 0 = 0 of course 0 > - sin 20
What is sin of -pi?
sin(-pi) = sin(-180) = 0 So the answer is 0
Sin and cosine values of 0?
sin 0 = 0 cos 0 = 1
Cos 0 sin 0 plus 1 equals 0?
No. sin(0) = 0 So cos(0)*sin(0) = 0 so the left hand side = 1
What is the sin of 2 pie?
sin(2*pi) - not pie - is the same as sin(0) = 0
What is the value of y sin θ when θ 0?
sin(0)=0, therefore ysin(0)=0
What is the sin of pi?
sin(pi) = 0
What is the lim of h if it equals 0 Sinxcosh plus cosxsinh minus sinx divided by h?
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 = ∞
Y equals 4 sin x for x equals pi?
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
What is the solution to sec plus tan equals cos over 1 plus sin?
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.
What circular functions are defined by sin 0?
Sin (0) isn't a function, it's a number.
What is the derivative of sin x?
f'[x] = lim(h->0) (f[x+h]-f[x])/h lim(h->0) (sin[x+h]-sin[x])/h By angle-addition formula, we have: lim(h->0) (sin[x]cos[h]+sin[h]cos[x]-sin[x])/h lim(h->0) (sin[x]cos[h]-sin[x])/h + lim(h->0) (sin[h]cos[x])/h sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h In a calculus class, it is shown that: lim(h->0) (cos[h]-1)/h = 0 and that lim(h->0) sin[h]/h is 1. So, sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h becomes sin[x]*0 + cos[x]*1 cos[x] So, if f[x] = sin[x], f'[x] = cos[x]
