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$\frac{1}{3}\cos^3 x \cos(2x)+\frac{1}{12}\sin(2x)(\sin(3x)+3\sin x)=\frac{1}{3} \cos x$

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?

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  • $\begingroup$ Ok I don't know why those white boxes are there. The expression is: (1/3)(cos^3(x)cos(2x))+ (1/12)(sin(2x)(sin(3x)+3sin(x))) = (1/3) cos (x) $\endgroup$ Jul 2, 2011 at 15:24
  • $\begingroup$ Yes, Mathematica has an "export to LaTeX" feature. $\endgroup$ Jul 2, 2011 at 15:59

2 Answers 2

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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...

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  • $\begingroup$ So, in your case, write everything as a rational function of $e^{ix}$. There is an algorithm to text equality of rational functions in one variable. $\endgroup$
    – GEdgar
    Jul 2, 2011 at 16:28
  • $\begingroup$ And if the identity only involves sines and cosines, then a Laurent polynomial in $e^{ix}$ will do, and that's even easier. $\endgroup$ Jul 2, 2011 at 16:30
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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.

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