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# Basic trig question watch

1. I feel embarrassed even asking this but I can't find it on the net... I know it'll be there somewhere but whatever I Google I'm either getting really basic stuff, or really advanced stuff...

What do I get if I multiply, say, cos2x by cos3x? Does it just stay as cos2xcos3x or can it be simplified further?

Thanks
2. I'm not sure what you mean by simplify here. You can't straight away say that e.g. cos2xcos3x=cos6x but you can reduce it into powers of cos (do you want this) using the angle sum formula

Note that cos(2x)=cos(x+x) and cos(3x)=cos(2x+x).

Have a go at reducing and post what you get.
3. cosA*cosB = 2 * cos((A+B)/2) * cos((A-B)/2)
4. (Original post by FieldLeftBlank)
cosA*cosB = 2 * cos((A+B)/2) * cos((A-B)/2)
Yes that's another way to simplify it if you know the formula. I'm not sure if it's in the formula booklet currently.
5. (Original post by notnek)
I'm not sure what you mean by simplify here. You can't straight away say that e.g. cos2xcos3x=cos6x but you can reduce it into powers of cos (do you want this) using the angle sum formula

Note that cos(2x)=cos(x+x) and cos(3x)=cos(2x+x).

Have a go at reducing and post what you get.
Ah right, that makes sense, and rings a bell, not sure if this will give me what I want

cos(2x) = cos(x)^2 - sin(x)^2
cos(3x) = cos(x)cos(2x) - sin(x)(sin2x)
cos(3x) = cos(x)(cos(x)^2 - sin(x)^2)...

Okay I can just see this going in circles Don't think this is the direction the question expects me to go in

I'll give this another shot, thanks for the help

EDIT: got it, I was missing something else that was really obvious... whoops

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