Deriving a Formula for cos 3a Using Only Cosine - Step by Step Guide

[rum2563]

Homework Statement

Derive a formula for cos 3a which would only involve cosine on the right side.

Homework Equations

cos2x = cos(x)^2 - sin(x)^2
cos2x = 2cos(x)^2 - 1
cos2x = 1-2sin(x)^2

The Attempt at a Solution

cos(3a) = cos(2a + a)
= cos2acosa + sin2acosa
= (2cos(a)^2 - 1)cosa - (2sinacosa)cosa
= 2cos(a)^3 - cosa - 2sina cos(a)^2


Am I on the right path? Or should I just continue to do keep doing this?

 

[rock.freak667]

cox(x+y)=cosxcosy-sinxsiny ...so your sign is wrong

 

[rum2563]

Sorry. My bad, It was such an obvious mistake.

Here it goes:

cos(3x) = cos(2x + x)

= cos2xcosx - sin2xsinx
= (2cos(x)^2 -1)cos(x) - (2sin(x)cos(x))cos(x)
= 2cos(x)^3 - cos(x) - 2sin(x)cos(x)^2
...

Ok. Now I am confused again. Please help me. Thanks.

 

[HallsofIvy]

That sign error was clearly a typo since you corrected it in the next line.

Your real problem is not the sign:

cos(3a) = cos(2a + a)
= cos2acosa + sin2acosa

No, that is cos(2a)cos(a)- sin(2a)sin(a)

= (2cos(a)^2 - 1)cosa - (2sinacosa)cosa
= 2cos(a)^3 - cosa - 2sina cos(a)^2

should be = (2cos^2(a)- 1)cos(a)- (2sin(a)cos(a))sin(a)
= 2cos^3(a)- cos(a)- 2sin^2(a)cos(a)

Now replace sin^2(a) with 1- cos^2(a).