Question

Cos(x+pi/3)+cos(x-pi/3)=cosx Have the show the statement and the rules? Thanks again

Answer 1

Recall that `cos(A +B) = cosAcosB - sinAsinB` and `cos(A-B) = cosAcosB + sinAsinB`.

`cosxcos(pi/3) - sinxsin(pi/3)+ cosxcos(pi/3) + sinxsin(pi/3) = cosx`

`2cosxcos(pi/3) = cosx`

`2cosx(1/2) = cosx`

`cosx= cosx`

`LHS = RHS`

As required.

Hopefully this helps!

Answer 2

Please see below.

Explanation:

.

`cos(x+pi/3)+cos(x-pi/3)=cosx`

We have a Sum to Product formula that gives us:

`cosalpha+cosbeta=2cos((alpha+beta)/2)cos((alpha-beta)/2)`

Using this formula, we get:

`cos(x+pi/3)+cos(x-pi/3)=2cosxcos(pi/3)=2cosx(1/2)=cosx`

Answer 3

See below.

Explanation:

Identities:

`color(red)bb(cos(A+B)=cosAcosB-sinAsinB)`

`color(red)bb(cos(A-B)=cosAcosB+sinAsinB)`

I'll do these in two parts, because of space:

`cos(x+pi/3)=cos(x)cos(pi/3)-sin(x)sin(pi/3)`

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

`cos(x-pi/3)=cos(x)cos(pi/3)+sin(x)sin(pi/3)`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =cos(x)(1/2)+sin(x)(sqrt(3)/2)`

`cos(x+pi/3)+cos(x-pi/3)`

`=cos(x)(1/2)-sin(x)(sqrt(3)/2)+cos(x)(1/2)+sin(x)(sqrt(3)/2)`

`cos(x)(1/2)-cancel(sin(x)(sqrt(3)/2))+cos(x)(1/2)+cancel(sin(x)(sqrt(3)/2))`

`cos(x)(1/2)+cos(x)(1/2)`

`cancel(2)cos(x)cancel((1/2))`

`cosx`

As required:

`LHS-=RHS`

Answer 4

Well, I don't know the rules on your Rule Menu, so I'm using mine.

LHS `=> cos(x+pi/3)+cos(x-pi/3)`

Applying, `color(red)(cos(A+B) = cosAcosB - sinAsinB` and `color(magenta)(cos(A-B) = cosAcosB + sinAsinB`

`color(white)(dd`

`=> color(red)(cosxcos(pi/3) -sinxsin(pi/3)) + color(magenta)(cosxcos (pi/3) +sinxsin(pi/3)`

`=> cancel2cosx(1/cancel2)` `color(white)(wwwwwwwww` `["as " cos(pi/3)=1/2]`

`=> cosx`