It is a seriously incorrect equation.
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
-x + 168 = 234-x = 66x = -66
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
x = 45 degrees sin(x) = cos(x) = 1/2 sqrt(2)
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 = ∞
x*1.05) = 168 so x = 168/1.05 => x = 160.
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
-x + 168 = 234-x = 66x = -66
84 x 2 = 168
136' x 168' = 22,848 square feet
x-intercept = -21y-intercept = 168
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
Sin[x] = Cos[x] + (1/3)
1/sin x = csc x
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0