y = 4 sin x + 3 cos x
dy/dx = 4 cos x + 3(−sin x) = 4 cos x − 3 sin x
y = 2 cos x − 5 sin x
dy/dx = 2(−sin x) − 5 cos x = −2 sin x − 5 cos x
y = 3 sin x
dy/dx = 3 cos x
d²y/dx² = 3(−sin x) = −3 sin x
The derivatives of cos x cycle with period 4:
cos x → −sin x → −cos x → sin x → cos x → …
10 = 4 × 2 + 2, so remainder 2.
d1: −sin x; d2: −cos x
d10/dx10[cos x] = −cos x
dy/dx = cos x − sin x
At x = π/4: cos(π/4) = √2/2, sin(π/4) = √2/2
Gradient = √2/2 − √2/2 = 0
This means x = π/4 is a stationary point of y = sin x + cos x.
f(x) = 2 cos x, f′(x) = −2 sin x
−2 sin x = 0 ⇒ sin x = 0
On [0, 2π]: x = 0, π, 2π
dy/dx = 2 cos x + 1 = 0 ⇒ cos x = −1/2
On [0, 2π]: x = 2π/3 and x = 4π/3
d²y/dx² = −2 sin x
At x = 2π/3: −2 sin(2π/3) = −2 × (√3/2) = −√3 < 0 ⇒ local max
y(2π/3) = 2 sin(2π/3) + 2π/3 = √3 + 2π/3
At x = 4π/3: −2 sin(4π/3) = −2 × (−√3/2) = √3 > 0 ⇒ local min
y(4π/3) = −√3 + 4π/3
At x = π/6: y = sin(π/6) = 1/2
dy/dx = cos x; at x = π/6: gradient = cos(π/6) = √3/2
Tangent: y − 1/2 = (√3/2)(x − π/6)
y = (√3/2)x − √3π/12 + 1/2
y = sin x
dy/dx = cos x
d²y/dx² = −sin x
d²y/dx² + y = −sin x + sin x = 0 ✓
This means sin x (and cos x) are solutions of the simple harmonic oscillator equation y′′ + y = 0, which models pendulums, springs, and oscillating electrical circuits.
x(t) = 3 sin t − cos t
v(t) = 3 cos t + sin t
a(t) = −3 sin t + cos t
v = 0: 3 cos t + sin t = 0
sin t = −3 cos t ⇒ tan t = −3
t = π − arctan(3) ≈ 1.893 rad or t = 2π − arctan(3) ≈ 5.034 rad