i think sin 200 is smaller than sin 0..
because sin 200= - sin 20..
sin 0 = 0
of course 0 > - sin 20
sin(0) is 0
cos(125) = cos(180 - 55) = cos(180)*cos(55) + sin(180)*sin(55) = -cos(55) since cos(180) = -1, and sin(180) = 0 So A = 55 degrees.
sin 0 = 0 cos 0 = 1
sin(2*pi) - not pie - is the same as sin(0) = 0
sin(pi) = 0
sin(0) is 0
sin(-pi) = sin(-180) = 0 So the answer is 0
smaller in amplitude: sin(x), -3/2 sin(x) cancel out to become -sin(x)/2, which has a smaller amplitude smaller wavelength: sin(x), sin(x), "combine" them by multiplying together. The wavelength is reduced by 2 If you are looking for an addition of waves that gets the smaller wavelength of a sine wave, here is the simplest one I can find. It is an infinite addition, and the result is sin(2x), a wave that has a smaller wavelength than the individual waves: sum from k=0 to infinity of sin(k*pi/2+z0)(2x-z0)k / k!
cos(125) = cos(180 - 55) = cos(180)*cos(55) + sin(180)*sin(55) = -cos(55) since cos(180) = -1, and sin(180) = 0 So A = 55 degrees.
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
60 milimeters are = with 0,06 meters (to convert gigger units into smaller units take a 0 on the right for every smaller unit and when you have no more 0s put a 0 on the left of the number that is not 0 and a , after the first 0)
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