Maths

f (x) = (cos (sin x) - cos x) / x? We need lim (x→0) f (x) = 1/k.
Using cos C - cos D = -2 sin (C+D)/2) sin (C-D)/2).
f (x) = -2 sin (sin x + x)/2) sin (sin x - x)/2) / x?
For small x, sin x ≈ x.
lim (x→0) f (x) = lim -2 * ( (sin x + x)/2 ) * ( (sin x - x)/2 ) / x?
Using series expansion: sin x = x - x³/3! + x? /5! - .
sin x + x = 2x - x³/6 + .
sin x - x = -x³/6 + x? /120 - .
f (x) ≈ -2 * ( (2x)/2 ) * ( (-x³/6)/2 ) / x?
≈ -2 * (x) * (-x³/12) / x?
≈ (2x? /12) / x? = 2/12 = 1/6.
So, 1/k = 1/6 ⇒ k = 6.

Kindly consider the following figure

Kindly consider the following figure

Kindly consider the following figure

Kindly consider the following figure

Kindly consider the following figure

We need to find the remainder of (2021)³? ² when divided by 17.
First, find the remainder of 2021 divided by 17.
2021 = 17 * 118 + 15.
So, 2021 ≡ 15 (mod 17).
Also, 15 ≡ -2 (mod 17).
So, (2021)³? ² ≡ (-2)³? ² (mod 17).
(-2)³? ² = 2³? ² = (2? )? ⋅ 2² = 16? ⋅ 4.
Since 16 ≡ -1 (mod 17),
16? ⋅ 4 ≡ (-1)? ⋅ 4 (mod 17).
≡ 1 ⋅ 4 (mod 17)
≡ 4 (mod 17).
The remainder is 4.

A = [0, sin α], [sin α, 0]
A² = A ⋅ A = [0, sin α], [sin α, 0] ⋅ [0, sin α], [sin α, 0]
= [00 + sinαsinα, 0sinα + sinα0], [sinα0 + 0sinα, sinαsinα + 00]
= [sin²α, 0], [0, sin²α] = (sin²α)I
A² - (1/2)I = [sin²α, 0], [0, sin²α] - [1/2, 0], [0, 1/2]
= [sin²α - 1/2, 0], [0, sin²α - 1/2]
det (A² - (1/2)I) = (sin²α - 1/2)² - 0 = 0
sin²α - 1/2 = 0
sin²α = 1/2
sin α = ±1/√2
A possible value for α is π/4.

Kindly consider the following figure