Question

Simplify `cos(2alpha+beta)-cos(2alpha-beta)`?

Answer 1

Read below.

Explanation:

First of all, remember that:
`cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)`
`cos(alpha-beta)=cos(alpha)cos(beta)+sin(alpha)sin(beta)`

Therefore, `cos(2alpha+beta)=cos(2alpha)cos(beta)-sin(2alpha)sin(beta)` Similarly,
`cos(2alpha-beta)=cos(2alpha)cos(beta)+sin(2alpha)sin(beta)`
Therefore,
`cos(2alpha+beta)-cos(2alpha-beta)=cos(2alpha)cos(beta)-sin(2alpha)sin(beta)-[cos(2alpha)cos(beta)+sin(2alpha)sin(beta)]`
=> `cos(2alpha)cos(beta)-sin(2alpha)sin(beta)-cos(2alpha)cos(beta)-sin(2alpha)sin(beta)`
Combine like terms
`cos(2alpha)cos(beta)-sin(2alpha)sin(beta)-cos(2alpha)cos(beta)-sin(2alpha)sin(beta)=>-2sin(2alpha)sin(beta)`
Now, remember that: `sin(alpha+beta)=sin(alpha)cos(beta)+sin(beta)cos(alpha)`

Therefore, `sin(2alpha)=sin(alpha)cos(alpha)+sin(alpha)cos(alpha)=>sin(alpha)cos(alpha)[2]`
So we have:
`-2sin(2alpha)sin(beta)=>-2[2]sin(alpha)cos(alpha)sin(beta)` We simplify this to get: `-2[2]sin(alpha)cos(alpha)sin(beta)=>-4sin(alpha)cos(alpha)sin(beta)`

Answer 2

Please see below.

Explanation:

As `cosC-cosD=-2sin((C+D)/2)sin((C-D)/2)` ....(A)

Let `C=2alpha+beta` and `D=2alpha-beta`

then `C+D=4alpha` and `C-D=2beta`

and then (A) becomes

`cos(2alpha+beta)-cos(2alpha-beta)=-2sin2alphasinbeta`

= `-2xx2sinalphacosalphaxxsinbeta`

= `-4sinalphacosalphasinbeta`