Simplifying trigonometric expressions, is there a unified theory? $\frac{1}{3}\cos^3 x \cos(2x)+\frac{1}{12}\sin(2x)(\sin(3x)+3\sin x)=\frac{1}{3} \cos x$
I got this as the result of a differential equation that I solved. The answer in the book is (1/3) cos(x), but after applying variation of parameters I got the expression on the left. To my delight Wolfram Alpha tells me that they are equal! (yay!!)
But, without cheating and using the computer, how would I ever know that? Nothing about my expression screams "simplify me" unless I'm missing something.
Perhaps I'd notice the graph looked like cosine if I happened to graph it.
I know many trig identities, but I have never heard of a formal procedure that always works to simplify. Is there such a thing? How would you approach this messy expression?
How can I get better at this important skill?
I have more problems to solve and it feels cheap to keep plugging my answers in to W.alpha to see if they are right.
Ps. Is there a widget to convert thing formatted for mathematical to latex and vice versa?
 A: Using complex exponentials gives a systematic way of verifying such identities. Start by substituting
$$
\cos t=\frac12(e^{it}+e^{-it}),\qquad \sin t=\frac1{2i}(e^{it}-e^{-it}),
$$
with $t=x$, $t=2x$ and $t=3x$ where appropriate. Remember that $(e^{ix})^n=e^{inx}$
and $e^{ix} e^{iy}=e^{i(x+y)}$. Expand the left hand side and compare...
A: Try to express everything on the $\text{l.h.s}$ in terms of $\cos$ and then see what happens. Use the following identities:


*

*$\sin{2x} = 2\cdot \sin{x} \cdot \cos{x}$

*$\sin{3x} = 3\sin{x} - 4 \sin^{3}{x}$

*$1-\sin^{2}{x}= \cos^{2}{x}$
So you have $$\frac{1}{3} \cos^{3}{x} \cdot \bigl[ 2 \cos^{2}{x}-1 \bigr]=\frac{2}{3}\cos^{5}{x} -\frac{1}{3}\cos^{3}{x} \qquad (\text{I})$$ and $$\frac{1}{12} \cdot 2\cdot \sin{x} \cos{x} \cdot \Bigl[ 6\sin{x} -4\sin^{3}{x}\Bigr]=\sin^{2}{x}\cdot \cos{x} -\frac{2}{3}\sin^{4}{x}\cdot\cos{x} \qquad (\text{II})$$
Now write $\sin^{2}{x}$ as $1-\cos^{2}{x}$ and $\sin^{4}{x}$ as $(1-\cos^{2}{x})^{2}$ and add equations $\text{I}$ and $\text{II}$ to get the final answer.
Added. Equation $\text{II}$ becomes $$(1-\cos^{2}{x}) \cdot \cos{x} - \frac{2}{3}(1-\cos^{2}{x})^{2}\cdot\cos{x}$$ $$ = \cos{x}-\cos^{3}{x} -\frac{2}{3} \cdot \Bigl[ \cos{x} - 2\cdot\cos^{3}{x} +\cos^{5}{x}\Bigr]$$ $$= \frac{1}{3}\cdot \cos{x} + \frac{1}{3}\cos^{3}{x} -\frac{2}{3}\cdot \cos^{5}{x} $$
The basic idea for simplifying such expressions is to observe the RHS and the LHS and then use double angle, half-angle formulas to simplify the equations. I don't think there are any such theoreies behind them.
