Can -3sin(x)sin(2x)+2cos(x)cos(3x) be simplified any further?

[Phenomniverse]

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Can this expression be simplified any further?
-3sin(x)sin(2x)+2cos(x)cos(3x)

 

[sinx]

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Can this expression be simplified any further?
-3sin(x)sin(2x)+2cos(x)cos(3x)

try trig identities, like double angle formulas, things like that

 

[mmm4444bot]

It can be factored:

2∙cos(x) ∙ [ a difference between a cosine term and a sine-squared term ]

I don't necessarily consider this a simplification, though. It depends on what you plan to do with it, afterwards. :cool:

 

[Phenomniverse]

Well I cam to this as a result of a differentiation problem, I'm not sure if its best to leave it as is or try to express it some other way. I'm not going to do anything with it afterwards.

 

[Dr.Peterson]

Well I cam to this as a result of a differentiation problem, I'm not sure if its best to leave it as is or try to express it some other way. I'm not going to do anything with it afterwards.

I'd probably leave it as is; possibly if you showed what you differentiated, we might have a different opinion based on its form. Some instructors actually tell you not to simplify in such a problem, because doing so only means more different forms for them to check. If this is a textbook problem, have you checked the back of the book to see whether they tend to simplify answers or not?

If I did want to do more with it, I would probably see whether using the angle-sum identity for cos(3x) = cos(x + 2x) would make it any easier to work with (I think it might), and possibly the double-angle identity would help too. But at that point my main interest would be to make it easier to do whatever came next, such as solving for zeroes.

Of course, it's always possible that we'll find that you differentiated incorrectly, which is another reason to show the original problem.

 

[Phenomniverse]

Here's the full working out of the differentiation (see attached pic)

 

[Dr.Peterson]

Here's the full working out of the differentiation (see attached pic)
View attachment 9995

As I suspected, you did this wrong. The two individual derivatives are incorrect.

You should be using the chain rule in each case, and you missed an important detail.

 

[Phenomniverse]

Ah I see, I shouldn't have changed the angles.
Now I have it like this (picture attached). Is that better? Does it need to be simplified further?

 

[Phenomniverse]

Or should the two terms be multiplied rather than added? My notes on the chain rule are a bit sketchy.

 

[MarkFL]

Or should the two terms be multiplied rather than added? My notes on the chain rule are a bit sketchy.

It appears to be correct now.

\(\displaystyle \displaystyle \frac{d}{dx}\sin(u(x))=\cos(u(x))\cdot\frac{d}{dx}u(x)\)

\(\displaystyle \displaystyle \frac{d}{dx}\cos(u(x))=-\sin(u(x))\cdot\frac{d}{dx}u(x)\)