What is difference between #2cosx# and #cos2x#?

As example:
#cos3x= cos2x*cosx# Why?
#2cos3x=?#

But how:
#2cos2x+2cos2x=?#
#2cos2x*2cos2x=?#

Could you explain and prove? How to avoid misunderstanding when solving complex equations? It's easy to just use the formula in simple cases

1 Answer
May 16, 2018

# 2cos x - cos 2x = -2cos^2x + 2 cos x - 1#

Explanation:

They're very different and your example is not correct.

# 2 cos x # is twice the cosine of angle #x#. It will be between #-2# and #2.#

# cos2x# is an abbreviation for #cos(2x).# It is the cosine of the angle #2x#, two times the angle #x#. The value of #cos 2x# will be between #-1# and #1.#

Before we think about double angles, let's remember the formula for the cosine of the sum of two angles:

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

We substitute #a=b=x.#

#cos 2x = cos(x + x) = cos x cos x - sin x sin x = cos ^2 x - sin ^2 x#

Since #cos^2 x + sin ^2x = 1# we can also write

#cos 2 x = cos^2x - (1- cos^2x) = 2 cos^2 x - 1#

and

#cos 2x = (1 - sin ^2 x) - sin^2 x = 1 - 2 sin ^2 x #

So there are (at least) three double angle formulas for cosine:

# cos 2x = cos ^2 x - sin ^2 x = 1 - 2 sin ^2 x = 2 cos^2 x - 1 #

The one with just cosine on the right is usually the most useful.

The example was #cos 3x.# Let's work out the triple angle formula for cosine.

#cos 3x = cos(x + 2x) = cos x cos 2 x - sin x sin 2x #

We need the double angle formula for sine, which is

#sin 2x = 2 sin x cos x#

#cos 3x = cos x (2 cos ^2 x -1) - sin x (2 sin x cos x) #

# cos 3x = cos x (2 cos ^2x - 1 - 2 sin^2 x ) #

# cos 3x = cos x (2 cos ^2x - 1 - 2 (1 - cos^2 x) ) #

# cos 3x = cos x (4 cos ^2x - 3 ) #

#cos 3x = 4 cos ^2 x - 3 cos x #

Let's literally answer the question:

# 2cos x - cos 2x = 2 cos x - (2 cos^2 x - 1) = -2cos^2x + 2 cos x - 1#