Question

Use trigonometric identities to find a general solution of the equation.

Answer 1

\[y''+9y = \sin^4x\]

So the trig ID's that I've been thinking about using are either \[\sin^2x = 1 - \cos^2 x\] or \[\sin^2 x = \frac{ 1 - \cos2x }{ 2 }\]
and of course these would be squared.

Answer 2

Best identity this and the cosine one \[\Large \sin x = \frac{e^{ix}-e^{-ix}}{2i}\] \[\large \sin^4 x = ( \frac{e^{ix}-e^{-ix}}{2i})^4 = \frac{1}{2^3} \frac{e^{i4x}-4e^{i2x}+6-4e^{-i2x}+e^{-4x}}{2}\] \[\large \frac{1}{8} \left( \frac{e^{i4x}+e^{-4x}}{2} - 4\frac{e^{i2x}+e^{-i2x} }{2} + \frac{6}{2}\right)\]\[\large \sin^4 x=\frac{1}{8} \left( \cos(4x)- 4 \cos(2x) +3 \right)\]

Answer 3

\[\color{red}{i^4 = (\sqrt{-1})^2\cdot (\sqrt{-1})^2}\]

Answer 4

\[\Large \cos x = \frac{e^{ix}+e^{-ix}}{2}\]