Question

Solve cos3x by product rule?

Answer 1

`cos 3x = 4 cos^3 x - 3 cos x`

Explanation:

I have no idea what the product rule is or how it applies.

There are a few different ways to get the triple angle formula for cosine. For example, it follows from De Moivre:

`cos 3x + i sin 3x = (cos x + i sin x)^3`

# = cos ^3 x + 3 i cos ^2 x sin x + 3 i^2 cos x sin^2 x + i^3 sin ^3 x #

`= (cos ^2 x - 3 cos x sin ^2 x) + i ( cos^2 x sin x - sin ^3 x)`

Equating real parts,

`cos 3x = \cos ^2 x - 3 \cos x sin ^2x = cos ^2 x - 3cos x(1 - cos ^2x )`

`cos 3x = 4 cos^3 x - 3 cos x`

The usual way to get the triple angle formula is from the sum angle formulas:

`cos(a+b)=cos a cos b - sin a sin b`

#sin(a+b)=sin a cos b + cos a sin b #

`cos(2a)=cos (a+a) = cos a cos a - sin a sin a = cos ^2 a - sin ^2 a = cos^2 a - (1-cos ^2 a) = 2 cos ^2 a - 1`

`sin(2a) = sin(a+a)=sin a cos a + cos a sin a = 2 sin a cos a`

`cos(3a) = cos(2a+a) = cos(2a)cos a - sin(2a) sin a`

`= (2 \cos ^2 a -1) cos a - 2 sin a cos a (sin a)`

`= 2 cos^3 a - cos a - 2 sin ^2 a cos a`

`= 2 cos^3 a - cos a - 2 (1- cos^2 a) cos a`

`cos(3a) = 4 cos ^3 a - 3 cos a`

Same answer.

Answer 2

First derivative: `-3sin(3x)`

Explanation:

I'm going to answer this question under the assumption you want the differentiation of `cos(3x)`.

`cos(3x)` is a composite function, not a product of two functions, so the Product Rule isn't applicable here.

`cos(3x)` is essentially made up of two functions

`f(x)=cosx`

`g(x)=3x`

Since we're dealing with a composite function, we can use the Chain Rule

`f'(g(x))*g'(x)`

We also know that

`f'(x)=-sinx` and `g'(x)=3`

Now, let's plug everything in. We get

`-sin(3x)*3`

which can be rewritten as

`-3sin(3x)`

Hope this helps!