Question about multiplying with trig
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Question about multiplying with trig

[From: ] [author: ] [Date: 11-05-18] [Hit: ]
....Im just unsure if you multiply a 2 in front of the cos as well, in which then.......
I have a homework equation that I must prove. The equation is:

(cos^2)2x - (sin^2)2x = cos4x

I started to break down the left side, and factored out a 2.
So my equation then looked like:

2(cos^2x - sin^2x) = "

Then I turned that into:

2(cos2x) = "

Now, if my work is correct, then I should continue to multiply the above.
When multiplying this, does it multiply out so the answer is:

2cos4x

or

cos4x (if it's the latter, then the problem has been proven)


....I'm just unsure if you multiply a 2 in front of the cos as well, in which then....well, in which then, I think I'd be stuck and not know what to do from there.
Think you can help me out?
thanks.

-
You cannot factor out the 2 because =2 is not a factor (multiplier)!

cos²(2x) – sin²(2x) has the 2 and the x inside the trig functions which are nonlinear, so you cannot "factor" anything out of them.

Yiou should know that cos 2θ = cos²θ – sin²θ is one of the basic double angle formulas. Just let θ = 2x and you get:

cos 2(2x) = cos² 2x – sin² 2x or more simply:

cos 4x = cos² 2x – sin² 2x

Remark: Hope this helps clarify why you cannot do what you tried to do. cos kθ ≠ k cos θ, nor can you do that for the sine. sin kθ ≠ k sin θ. You cannot take a constant from inside a trig function (or ANY other nonlinear function, like square roots for example) like you do for multiplication. In higher math we say that these functions are non-homogeneous. In fact the *only* function you can do this for is f(x) = cx where c is a constant.
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