Notes: What is cos(x+2pi)+cos(x-2pi)?

Saturday, March 29, 2014

We'll transform the sum into a product using the
formula:

$\displaystyle{\cos{{a}}}+{\cos{{b}}}=$
2cos[(a+b)/2]*cos[(a-b)/2]

Let $\displaystyle{a}={x}+{2}\pi{\quad\text{and}\quad}{b}={x}-$
2pi

$\displaystyle{a}+{b}={x}+{2}\pi+{x}-{2}\pi=$
2x

$\displaystyle{a}-{b}={x}+{2}\pi-{x}+$
2pi

$\displaystyle{a}-{b}={4}\pi$

$\displaystyle{\cos{{\left({x}+\right.}}}$
2pi) + cos (x - 2pi) = 2 cos (2x/2)*cos (4pi/2)

$\displaystyle{\cos{{\left({x}+\right.}}}$
2pi) + cos (x - 2pi) = 2 cos x*cos 2pi

$\displaystyle{B}{u}{t}{\cos{{2}}}\pi=$
1

$\displaystyle{\cos{{\left({x}+{2}\pi\right)}}}+{\cos{{\left({x}-{2}\pi\right)}}}={2}{\cos{}}$
x

The sum can be calculated using the following
identities:

$\displaystyle{\cos{{\left({x}+{2}\pi\right)}}}={\cos{{x}}}\cdot{\cos{{2}}}\pi-{\sin{{x}}}\cdot{\sin{}}$
2pi

Since $\displaystyle{\sin{{2}}}\pi={0}$ , the product $\displaystyle{\sin{{x}}}\cdot{\sin{{2}}}\pi$ will
be 0.

$\displaystyle{\cos{{\left({x}+{2}\pi\right)}}}={\cos{}}$
x

$\displaystyle{\cos{{\left({x}-{2}\pi\right)}}}={\cos{{x}}}\cdot{\cos{{2}}}\pi+{\sin{{x}}}\cdot{\sin{}}$
2pi

$\displaystyle{\cos{{\left({x}-{2}\pi\right)}}}={\cos{}}$
x

$\displaystyle{\cos{{\left({x}+{2}\pi\right)}}}+{\cos{{\left({x}-{2}\pi\right)}}}={\cos{{x}}}+{\cos{{x}}}={2}{\cos{}}$
x

Therefore, the requested sum is $\displaystyle{\cos{{\left({x}+\right.}}}$
2pi) + cos (x - 2pi) = 2 cos x.

Notes