Diketahui tan 3x= n , maka nilai sin 12 x

trigonometri

tan (3x) = n = y/x
y = n
x = 1
r² = x² + y² = n² + 1
r = √(n²+1)
.
sin (3x) = y/r = n/√(n²+1)
cos (3x) = x/r = 1/ √(n²+1)
sin 3x. cos 3x = n/(n²+1)
sin² (3x) = n²/(n²+1)
.
misal (3x) = A
sin (12 x) = sin  4 (3x) = sin 4A = sin 2 (2A)
= 2 sin 2A cos 2A
= 2 ( 2 sin A cos A) (1 - 2 sin² A)
= 2 (2 sin 3x cos 3x) (1 - 2 sin² 3x)
= (4 sin3x cos 3x) (1 - 2 (sin 3x)²)
= (4.n/(n²+1)) (1 - 2 (n²/(n²+1))
= 4n/(n²+1) - 8n³/(n²+1)
= (4n -8n³)/(n²+1)