Using Sum/diff Identities to prove the identity

[yeunju]

Problem:
cos x + cos 2x + cos 3x = (2cosx + 1) cos 2x

EDIT: sorry for not being clear
Sum/difference identities:
a. cosine of sum or difference= cos (u +/- v) = cos(u)cos(v) -/+ sin(u)sin(v) : here the addition and subtraction would switch
just another example: of these types, but wouldn't be used for the problem.
b. sine of sum or difference= sin(u +/- v)= sin(u)cos(v) +/- cos (u)sin(v): here the addition and subtraction would be the same..

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My attempt:
Starting on left.
cos (x) + cos (x+x) + cos (2x+x)
cos (x) + cosxcosx - sinxsinx + cos2xcosx - sin2xsinx
cos (x) + (cosx)^2 - (sinx)^2 +cos(x+x)cosx - sin(x+x)sinx :used cosine of sum/diff, again
cos(x) + (cosx)^2 - (sinx)^2 + (cosxcosx - sinxsinx) cosx - (sinxcosx + sinxcosx) sinx :Thought the bold would work for pythagorean identity, but it can't because its subtracting sinx

then it gets messier =w=;; Currently i'm trying from the right because i gave up on the left.

 

[royhaas]

Re: Using Sum Identities to prove the identity

By "sum identities", do you mean results of things like $\displaystyle \cos(\frac{A+B}{2})+\cos(\frac{A-B}{2})$ ?

 

[soroban]

Hello, yeunju!

Here's a back-door approach . . .

$\displaystyle \text{Use sum/diff identities to prove: }\:\cos x + \cos 2x + \cos 3x \:=\: (2\cos x + 1)\cos 2x$

$\displaystyle \text{Note that: }\;\begin{array}{cccccc}\cos x &=& \cos(2x-x) &=& \cos2x\cos x - \sin2x\sin x &[1] \\ \cos3x &=& \cos(2x+x) &=& \cos2x\cos x + \sin2x\sin x & [2] \end{array}$

. . $\displaystyle \text{Add [1] and [2]: }\;\cos x + \cos3x \;=\;2\cos2x\cos x$ .**


$\displaystyle \text{The problem becomes:}$

. . $\displaystyle (\cos x + \cos3x) + \cos2x \;=\;2\cos2x\cos x + \cos2x \;=\;(2\cos x+1)\cos2x$


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** .This is one of the Sum-to-Product Identites.

. . . . $\displaystyle \begin{array}{ccc} \sin A + \sin B &=&2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \\ \sin A - \sin B &=& 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \\ \\ \cos A + \cos B &=& 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \\ \cos A - \cos B &=& \text{-}2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \end{array}$