sin(0)=0, therefore ysin(0)=0
lim(x->0) of sin(x)^2/x we use L'Hospital's Rule and derive the top and the bottomd/dx(sin(x)^2/x)=2*sin(x)*cos(x)/1lim(x->0) of 2*sin(x)*cos(x)=2*0*1=0
you need to explain it better but with what i got i if sin x equals b obviously sin 2x is double of b hence sin 2x is more than b. {Not obvious at all, actually. And the above is false. Sin(Pi/2) = 1 and Sin(Pi)=0. But clearly 2 is not greater than 0. Contradiction.} Obtuse means Pi/2 < x < Pi So, sin(x) = b means b>0, because sin(y) > 0 if 0<y<Pi sin(2x)=2sin(x)cos(x) cos(x) < 0 because cos(Pi/2)= 0 and the derivative is negative there. Hence, sin(2x) = 2 sin(x) cos(x) = 2*b*(-K), where K is a positive constant Since b>0, -2Kb < b
Let g(x) = interval [0, x] of sin t dt, and f(t) = sin t. Since f(t) is a continuous function, the part one of the Fundamental Theorem of Calculus gives, g'(x) = sin x = f(x) (the original function). If you are interested in the interval [x, 0] of sin t dt, then just put a minus sign in front of the integral and interchange places of 0 and x. So that, g(x) = interval [x, 0] of sin t dt = -{ interval [0, x] of sin t dt}, then g'(x) = - sin x.
the value of sin(x) lies between -1 to +1. the approx value of sin(x)/x = 1 when x tends to 0 & sin(x)/x = 0 when x tends to infinity.
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
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.
1.4 Classification Of FunctionsAnalytically represented functions are either Elementary or Non-elementary.The basic elementary functions are :1) Power function :y = xm , m ÎR2) Exponential function :y = ax , a > 0 but a ¹ 13) Logarithmic function :y = log ax , a > 0, a ¹ 1 and x > 04) Trigonometric functions :y = sin x, y = cos x, y = tan x,y = csc x, y = sec x and y = cot x5) Inverse trigonometric functionsy = sin-1 x, y = cos-1x, y = tan-1x,OR y = cot-1x, y = cosec-1x, y = sec-1x.y = arc sin x, y = arc cos x, y = arc tan xy = arc cot x, y = arc csc x and y = arc sec x
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