Nilai maksimum dari fungsi trigonometri : 1. y = 2 + 3 cos²x 2. y = 1 - sin²x

y = 2 + 3cos² x
y maksimum jika y' = 0
y' = -3sin 2x = 0
→ sin 2x = 0
→ x = (nπ)/2, n = 0, 1, 2, ...
Ambil x = 0°, maka
y = 2 + 3cos² (0°) = 5

y = 1 - sin² x
y maksimum jika y' = 0
y' = -sin 2x = 0
→ x = (nπ)/2, n = 0, 1, 2, ...
Ambil x = 0°, maka
y = 1 - sin² (0°) = 1