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How do you prove sin x tan x equals cos 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.
If Sin equals x and Cos equals y then x squared equals what function of y?
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
If sin x - cos x equals 1 over 3 what is sin x?
Sin[x] = Cos[x] + (1/3)
Sin x Tan x equals Sin x?
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
How do you solve the following identity sec x - cos x equals sin x tan x?
sec x - cos x = (sin x)(tan x) 1/cos x - cos x = Cofunction Identity, sec x = 1/cos x. (1-cos^2 x)/cos x = Subtract the fractions. (sin^2 x)/cos x = Pythagorean Identity, 1-cos^2 x = sin^2 x. sin x (sin x)/(cos x) = Factor out sin x. (sin x)(tan x) = (sin x)(tan x) Cofunction Identity, (sin x)/(cos x) = tan x.
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 = ∞
Verify the identity sinx cotx - cosx divided by tanx equals 0?
(sin(x)cot(x) - cos(x))/tan(x)(Multiply by tan(x)/tan(x))sin(x) - cos(x)tan(x)(tan(x) = sin(x)/cos(x))sinx - cos(x)(sin(x)/cos(x))(cos(x) cancels out)sin(x) - sin(x)0
How do you solve sin 2x equals sin x over 2?
they do have calculators for these questions you knowsin 2x = (sin x)/22 sin x cos x - (1/2)sin x = 02 sin x(cos x - 1/4) = 02 sin x = 0 or cos x - 1/4 = 0sin x = 0 or cos x = 1/4in the interval [0, 360)sin x = 0, when x = 0, 180cos x = 1/4, when x = 75.52, 284.48Check:
What is cosx-2sinxcosx equals 0?
cos x - 2 sin x cos x = 0 -> cos x (1 - 2 sin x) = 0 => cos x = 0 or 1 - 2 sin x = 0 cos x = 0: x = π/2 + kπ 1 - 2 sin x = 0: sin x = 1/2 -> x = π/6 + 2kπ or 5/6π + 2kπ Thus x = π/2 + kπ; x = π/6 + 2kπ; x = 5/6π + 2kπ solve the original equation.
How do you prove sin x tan x equals cos 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.
How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(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
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]
If Sin equals x and Cos equals y then x squared equals what function of y?
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
If sin x - cos x equals 1 over 3 what is sin x?
Sin[x] = Cos[x] + (1/3)
How do you factor sin squared times x plus cos2x -cosx equals 0?
2
Sin x Tan x equals Sin x?
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
When does cos x equal -sin x?
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
