How to simplify a trig identity?

Question

(1/3)(cos^3(x)cos(2x))+(1/12)(sin(2x)(sin(3x)+3sin(x))) = (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.

Answer 1

(1/3) * cos(x)^3 * cos(2x) + (1/12) * sin(2x) * (sin(3x) + 3sin(x))) =>

(1/3) * cos(x)^3 * (cos(x)^2 - sin(x)^2) + (1/12) * 2 * sin(x) * cos(x) * (sin(2x)cos(x) + sin(x)cos(2x) + 3sin(x)) =>

(1/3) * cos(x)^3 * (2cos(x)^2 - 1) + (1/6) * sin(x) * cos(x) * (2sin(x)cos(x)cos(x) + sin(x) * (cos(x)^2 - sin(x)^2) + 3sin(x)) =>

(1/3) * cos(x)^3 * (2cos(x)^2 - 1) + (1/6) * sin(x) * cos(x) * (2sin(x) * cos(x)^2 + sin(x) * cos(x)^2 - sin(x)^3 + 3sin(x)) =>

(1/3) * cos(x)^3 * (2cos(x)^2 - 1) + (1/6) * sin(x) * cos(x) * (3sin(x) * cos(x)^2 - sin(x)^3 + 3sin(x)) =>

(1/3) * cos(x)^3 * (2cos(x)^2 - 1) + (1/6) * sin(x)^2 * cos(x) * (3cos(x)^2 - sin(x)^2 + 3) =>

(1/3) * cos(x)^3 * (2cos(x)^2 - 1) + (1/6) * sin(x)^2 * cos(x) * (3 - 3sin(x)^2 - sin(x)^2 + 3) =>

(1/3) * cos(x)^3 * (2cos(x)^2 - 1) + (1/6) * sin(x)^2 * cos(x) * (6 - 4sin(x)^2) =>

(1/3) * cos(x)^3 * (2cos(x)^2 - 1) + (1/6) * 2 * sin(x)^2 * cos(x) * (3 - 2sin(x)^2) =>

(1/3) * cos(x)^3 * (2cos(x)^2 - 1) + (1/3) * sin(x)^2 * cos(x) * (3 - 2sin(x)^2) =>

(1/3) * (cos(x)^3 * (2cos(x)^2 - 1) + sin(x)^2 * cos(x) * (3 - 2sin(x)^2)) =>

(1/3) * (2cos(x)^5 - cos(x)^3 + cos(x) * (1 - cos(x)^2) * (3 - 2 * (1 - cos(x)^2))) =>



For the sake of clarity, let's just say that cos(x) = t



(1/3) * (2t^5 - t^3 + t * (1 - t^2) * (3 - 2 + 2t^2)) =>

(1/3) * (2t^5 - t^3 + t * (1 - t^2) * (1 + 2t^2)) =>

(1/3) * (2t^5 - t^3 + t * (1 - t^2 + 2t^2 - 2t^4)) =>

(1/3) * (2t^5 - t^3 + t * (1 + t^2 - 2t^4)) =>

(1/3) * (2t^5 - t^3 + t + t^3 - 2t^5) =>

(1/3) * t =>

(1/3) * cos(x)

Answer 2

There is no one hard and fast rule to simplify all trig expressions, but the fact that you have double angles and triple angles of trig functions along with single values of x is a pretty good clue that if you get rid of all of your double and triple angles things might simplify.



Hence start off with replacing those trig ids as follows:



cos(2x) = cos²x - sin²x

sin(2x) = 2sinxcosx

sin(3x)=sin(2x)cosx + cos(2x)sinx = (2sinxcosx)cosx + (cos²x-sin²x)sinx = 2sinxcos²x+sinxcos²x-sin³x



After doing all of these replacements, you can multiply through and simplify and it will turn out that everything cancels except the (1/3)cosx

Answer 3

So that first part expands and you get cosx + (cosx * tan^2x) Expand out tan^2x cosx + (cosx * (sin^2x/cos^2x)) Multiply and simplify the terms in the parentheses cosx + (sin^2x/cosx) Get both terms over the same term so that they can add easily (put cosx over one, multiply the sin^2x by the denom of the cosx term and multiply the cosx by the sin^2x denom and then multiply the denoms) (cos^2x/cosx) + (sin^2x/cosx) Combine the terms because they are under the same fraction (cos^2x+sin^2x)/cosx cos^2x+sin^2x = 1 in trigonometric identities so it reduces to 1/cosx Which also equals secx PS this is exactly what the guy above me posted but minus the 1+tan^2x shortcut. There are many ways to do these identities and he just did the easier one, but I had forgotten the shortcut and took the easy way. Both are right.