Q: Why does cos x - sin x 0?

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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]

(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

sin(x) = x - x3/3! + x5/5! - x7/7! + ... and cos(x) = 1 - x2/2! + x4/4! - x6/6! + ... where x is the angle measured in radians. Then tan(x) = sin(x)/cos(x) where cos(x) is not 0 cosec(x) = 1/sin(x) where sin(x) is not 0 sec(x) = 1/cos(x) where cos(x) is not 0 and cot(x) = cos(x)/sin(x) where sin(x) is not 0

Cos(x) = Sin(2x) Using angle-addition, we have Sin(a+b) = Sin(a)Cos(b) + Sin(b)Cos(a). From that, we see Sin(2x) = Sin(x)Cos(x)+Sin(x)Cos(x) = 2Sin(x)Cos(x) Cos(x) = 2Sin(x)Cos(x) If Cos(x) = 0, then the two sides are equal. This occurs at x= Pi/2 + nPi, where n is an integer and Pi is approximately 3.14. If Cos(x) doesn't equal 0, then we can divide it out. Then, 1 = 2 Sin(x) , or 1/2 = Sin(x) This occurs when x = Pi/6 or 5Pi/6, plus or minus any multiples of 2 Pi.

(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True

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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 = ∞

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]

(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

sin(x) = x - x3/3! + x5/5! - x7/7! + ... and cos(x) = 1 - x2/2! + x4/4! - x6/6! + ... where x is the angle measured in radians. Then tan(x) = sin(x)/cos(x) where cos(x) is not 0 cosec(x) = 1/sin(x) where sin(x) is not 0 sec(x) = 1/cos(x) where cos(x) is not 0 and cot(x) = cos(x)/sin(x) where sin(x) is not 0

tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).

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:

(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

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.

if tan x = cos x then sin x / cos x = cos x => sin x = cos x cos x => sin x = cos2 x => sin x = 1 - sin2x => sin2x + sin x - 1 = 0 Using the quadratic formula => 1. sin x = 0.61803398874989484820458683436564 => x = sin-1 (0.61803398874989484820458683436564) or => 2. sin x = -1.6180339887498948482045868343656 => x = sin-1 (-1.6180339887498948482045868343656)

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Cos(x) = Sin(2x) Using angle-addition, we have Sin(a+b) = Sin(a)Cos(b) + Sin(b)Cos(a). From that, we see Sin(2x) = Sin(x)Cos(x)+Sin(x)Cos(x) = 2Sin(x)Cos(x) Cos(x) = 2Sin(x)Cos(x) If Cos(x) = 0, then the two sides are equal. This occurs at x= Pi/2 + nPi, where n is an integer and Pi is approximately 3.14. If Cos(x) doesn't equal 0, then we can divide it out. Then, 1 = 2 Sin(x) , or 1/2 = Sin(x) This occurs when x = Pi/6 or 5Pi/6, plus or minus any multiples of 2 Pi.

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