3 years ago
(a) Write
cos 3x + i sin 3x = e^(3ix) = (cos x + i sin x)^3
by Euler's formula, expand, and equate real and imag inary parts to derive the identities
cos^(3) x = (3/4) cos x + (1/4) cos 3x
sin^(3) x = (3/4) sin x - (1/4) sin 3x
(b) Use the result of part (a) to find a general solution of
y"" + 4y = cos^(3) x.
1 Answers
Best Answer
cxnguyen16 Staff answered 3 years ago
cxnguyen16 Staff answered 3 years ago
a) Given
The equation is cos 3x+i sin 3x = e^(3) i x = (cos 3x+i sin 3x)^(3)
Expand the equation
cos 3x + i sin 3x = cos^(3) x + 3 i cos^(2) x sin x - 3 cos x sin^(2) x - i sin^(3) x
= cos^(3) x + 3 i (1 - sin^(2) x) sin x - 3 cos x (1 - cos^(2) x) - i sin^(3) x
= cos^(3) x + 3 i sin x - 3 i sin^(3) x - 3 cos x + 3 cos^(3) x - i sin^(3) x
= 4 cos^(3) x + 3 i sin x - 3 cos x - 4 i sin^(3) x
= (4 cos^(3) x - 3 cos x) + i (3 sin x - 4 sin^(3) x)
Equate the real part:
cos 3x = 4 cos^(3) x - 3 cos x
4 cos^(3) x = cos 3 x + 3 cos x
cos^(3) x = (1/4) cos 3x + (3/4) cos x
Equate the imaginary part:
sin 3 x = 3 sin x - 4 sin^(3) x
4 sin^(3) x = 3 sin x - sin 3 x
sin^(3) x = (3/4) sin x - (1/4) sin 3 x
Hence proved
The equation is cos 3x+i sin 3x = e^(3) i x = (cos 3x+i sin 3x)^(3)
Expand the equation
cos 3x + i sin 3x = cos^(3) x + 3 i cos^(2) x sin x - 3 cos x sin^(2) x - i sin^(3) x
= cos^(3) x + 3 i (1 - sin^(2) x) sin x - 3 cos x (1 - cos^(2) x) - i sin^(3) x
= cos^(3) x + 3 i sin x - 3 i sin^(3) x - 3 cos x + 3 cos^(3) x - i sin^(3) x
= 4 cos^(3) x + 3 i sin x - 3 cos x - 4 i sin^(3) x
= (4 cos^(3) x - 3 cos x) + i (3 sin x - 4 sin^(3) x)
Equate the real part:
cos 3x = 4 cos^(3) x - 3 cos x
4 cos^(3) x = cos 3 x + 3 cos x
cos^(3) x = (1/4) cos 3x + (3/4) cos x
Equate the imaginary part:
sin 3 x = 3 sin x - 4 sin^(3) x
4 sin^(3) x = 3 sin x - sin 3 x
sin^(3) x = (3/4) sin x - (1/4) sin 3 x
Hence proved