Using the identity `cos(A+B) = cosAcosB - sinAsinB`, how do you prove that `1/4cos(3A) = cos^3A - 3/4cosA`?

I'm going to work from the right hand side. For `cos3A`, we know that:

`color(green)(cos3A = cos(A + 2A) = cosAcolor(highlight)(cos2A) - sinAcolor(highlight)(sin2A)).`

Next, recall that:

• `color(highlight)(sin2A) = sinAcosA + cosAsinA = color(highlight)(2sinAcosA)`
• `color(highlight)(cos2A) = cosAcosA - sinAsinA = color(highlight)(cos^2A - sin^2A)`

So, we now have:

`cos3A = cosA(color(highlight)(cos^2A - sin^2A)) - sinA(color(highlight)(2sinAcosA))`

`= cos^3A - sin^2AcosA - 2sin^2AcosA`

`= cos^3A - 3color(red)(sin^2)AcosA`

What I'm shooting for at this point is the result you get on the left hand side when you multiply the right hand side by `4`: `color(purple)(4cos^3A - 3cosA)`. That requires all functions to be `cos` here.

So, we have to reference the identity that you definitely know: `color(red)(sin^2A + cos^2A = 1)`. This gives us:

`= cos^3A - 3(color(red)(1 - cos^2A))cosA`

`= cos^3A - 3(cosA - cos^3A)`

`= cos^3A - 3cosA + 3cos^3A`

`= color(purple)(4cos^3A - 3cosA)`

Almost there! Finally, since this was equal to `cos3A`, we have to divide by `4` to get:

`1/4cos3A = 1/4*[4cos^3A - 3cosA]`

`=> color(blue)(1/4cos3A = cos^3A - 3/4cosA)`

General Discussion

The identity to be assumed is

`cos(A+B)=cosAcosB-sinAsinB...(1)`

As this is valid for all real values of A and B , we can put convenient values to get different identities as follows.

Method - I( A Tricky one)

Putting `B= -B` in equation (1)

`cos(A-B)=cosAcos(-B)-sinAsin(-B)`

`=>cos(A-B)=cosAcosB+sinAsinB…(1.1)`

Adding (1) and (1.1) we get

`cos(A+B)+cos(A-B)=2cosAcosB…(1.2)`

Now Putting `B=2A` in (1.2)

`cos(A+2A)+cos(A-2A)=2cosAcos2A`

`cos3A+cosA=2cosAcos2A`

`cos3A=cosA(2cos2A-1)…(1.3)`

Putting `B=A` in (1.2)

`cos(A+A)+cos(A-A)=2cosAcosA`

`=>cos2A+cos0=2cos^2A`

`cos2A=2cos^2A-1…(1.4)`

Finally combining (1.4) with (1.3) we get

`cos3A=cosA(2(2cos^2A-1)-1)`

`=>cos3A=4cos^3A-3cosA`

Re-arrenging and deviding bothsides by 4
`cos^3A-3/4cosA=1/4cos3A`
Proved

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method-II (A long one)

Putting `B=A` in equation(1)

`cos(A+A)=cosAcosA-sinAsinA`

`cos2A=cos^2A-sin^2A....(2)`

Putting`B=90+A`in equation(1)

`cos(A+90+A)=cosAcos(90+A)-sinAsin(90+A)`

`-sin2A=-sinAcosA-sinAcosA`

`=>sin2A=2sinAcosA...(3)`

Putting `B=-A` in equation(1)

`cos(A-A)=cosAcos(-A)-sinAsin(-A)`

`=>cos^2A+sin^2A=1...(4)`

Now finally putting `B=2A` in equation (1)

`cos(A+2A)=cosAcos2A-sinAsin2A`

inserting the relation (3)

`cos3A=cosAcos2A-2sinAsinAcosA`

`=cosA(cos2A-2sin^2A)`

Inserting relation (2)

`cos3A=cosA(cos^2A-sin^2A-2sin^2A)`

`=cosA(cos^2A-3sin^2A)`

`=cosA(4cos^2A-3cos^2A-3sin^2A)`

`=cosA(4cos^2A-3(cos^2A+sin^2A)`

`=cosA(4cos^2A-3*1)" " relation(4)`

`cos3A=4cos^3A-3cosA`

Re-arrenging and deviding bothsides by 4

`:.cos^3A-3/4cosA=1/4cos3A`

Proved

RHS is
`1/4cos(3A) = 1/4cos(2A+A)`
Using the given expression
`=> 1/4(cos (2A)cos(A) - sin(2A)sinA)`

1. Substituting the known expression for `sin (2A)=2sinA cos A`
2. Using given expression again for
   `cos(2A)=cos (A+A)=cos^2A-sin^2A`
3. and writing all terms in `"cosine"` using `sin^2A+cos^2A=1` we obtain
   `1/4((2cos^2A-1)cosA - 2sin^2AcosA)`

`=> 1/4(2cos^3A - cosA - 2(1-cos^2A)cosA)`
`= 1/4(2cos^3A - cosA - 2cosA + 2cos^3A)`
`=>1/4( 4cos^3A - 3cosA)`
`=> cos^3A - 3/4 cosA=`LHS