Question 1068643
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cos^3(A)×cos(3A)+sin^3(A)×sin(3A)=cos^3 (2A)
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Unfortunately, the person who submitted this post didn't say what he/she wanted.

So, I will formulate it instead of him/her: "Prove an identity . . . "


<pre>
The keys are these two formulas:

    {{{cos^3(A)}}}  = {{{(1/4)*cos(3A) + (3/4)*cos(A)}}},        (1)   and

    {{{sin^3(A)}}}  = {{{(1/4)*sin(3A) - (3/4)*sin(A)}}}.        (2)

(see the lesson <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Powers-of-trigonometric-functions.lesson>Powers of trigonometric functions</A> in this site).  When applying them, you will get

cos^3(A)*cos(3A) =  {{{(1/4)*cos(3A) + (3/4)*cos(A))*cos(3A)}}},   (3)  and

sin^3(A)*sin(3A) = {{{(-1/4)*sin(3A) - (3/4)*sin(A))*sin(3A)}}}.   (4)


So, adding and expanding (3) and (4), you will get

  {{{cos^3(A)*cos(3A)+sin^3(A)*sin(3A)}}} = 

= {{{(1/4)*cos^2(3A) + (3/4)*cos(A)*cos(3A) - (1/4)*sin^2(3A) + (3/4)*sin(A)*cos(3A)}}} = 

= [{{{(1/4)*cos^2(3A) - (1/4)*sin^2(3A)}}}] + [{{{(3/4)*cos(A)*cos(3A) + (3/4)*sin(A)*cos(3A))}}}] = 

    For the first  bracket  [ . . ]  apply the formula cos(2x) = . . .   
    For the second bracket  [ . . ]  apply the formula cos(x-y) = . . .   You will get

= {{{(1/4)*cos(6A) + (3/4)*cos(A-3A)}}} = {{{(1/4)*cos(6A) + (3/4)*cos(2A)}}} = 

    Now apply again the formula (1). You will get

= {{{cos^3(2A)}}}.
</pre>

QED.  Proved and solved.