Jawab:
= ¼ · (4 - π)√2 satuan

Penjelasan:
g dan h garis lurus yg menyinggung
y = sin(x) + cos(x)  {0 ≤ x ≤ 2π}

g dan h tegak lurus terhadap y = x
y = 1x (sesuai dengan y = mx + k, maka)
m = 1
m × mh = -1
1 × mh = -1
-->  mh = -1

y = sin(x) + cos(x)
-->  y' = cos(x) - sin(x)

mh = cos(x) - sin(x)
-1 = cos(x) - sin(x)
cos(x) = 1 - sin(x)
cos²(x) = (1 - sin(x))²
cos²(x) = 1 - 2sin(x) + sin²(x)
cos²(x) = 1 - 2sin(x) + sin²(x)
∵ sin²(x) + cos²(x) = 1 ∴
1 - sin²(x) = 1 - 2sin(x) + sin²(x)
1 - sin²(x) - sin²(x) = 1 - 2sin(x)
1 - 2sin²(x) = 1 - 2sin(x)
2sin²(x) = 2sin(x)
2sin²(x) - 2sin(x) = 0
Subs u dengan sin(x) maka
2u² - 2u = 0
2u(u-1) = 0
u = 0, u = 1
sin(x) = 0, sin(x) = 1
x = 0, π    x = π/2

Cek nyinggung atau enggak
-1 = cos(x) - sin(x)
-1 = cos(0) - sin(0)
-1 = 1 - 0 (tidak memenuhi)

-1 = cos(x) - sin(x)
-1 = cos(π) - sin(π)
-1 = -1 - 0 (memenuhi)

-1 = cos(x) - sin(x)
-1 = cos(π/2) - sin(π/2)
-1 = 0 - 1 (memenuhi)

Artinya
x = {π, π/2}

Garis g
Untuk x₁ = π
y₁ = sin(x₁) + cos(x₁)
y₁ = sin(π) + cos(π)
y₁ = 0 + (-1)
y₁ = -1
Persamaan garis g
y - y₁ = (mh)(x - x₁)
y - (-1) = -1(x - π)
y + 1 = -x + π
-->  x + y + 1 - π = 0

Garis h
Untuk x₁ = π/2
y₁ = sin(x₁) + cos(x₁)
y₁ = sin(π/2) + cos(π/2)
y₁ = 1 + 0
y₁ = 1
Persamaan garis h
y - y₁ = (mh)(x - x₁)
y - 1 = -1(x - (π/2))
y - 1 = -x + (π/2)
y = -x + (π/2) + 1
-->  x + y - 1 - (π/2) = 0

Gabungkan
g   -->   x + y + 1 - π = 0
A = 1, B = 1, C = 1 - π
h   -->   x + y - 1 - (π/2) = 0
..              D = - 1 - (π/2)

Jarak garis g dan h
= (| C - D |) / √(A²+B²)
= (|1 - π - (- 1 - (π/2))|) / √(1²+1²)
= (|1 - π + 1 + (π/2)|) / √2
= (|2 - π + (π/2)|) / √2
= (|(4/2) - (2π/2) + (π/2))|) / √2
= (|4 - 2π + π|) / (2√2)
= (|4 - π|) / (2√2)
= (4 - π)√2 / (2√2√2)
= (4 - π)√2 / (2(2))
= (4 - π)√2 / 4 satuan
= ¼ · (4 - π)√2 satuan

(xcvi)