Question 1031286
.
He all,
can you please help me verifying this identity step by step?:

sin(3α)=3sin(α)-4sin^3(α)

Thank you very much in advance 
RB
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<pre>
Do you know this formula:  {{{sin(alpha + beta)}}} = {{{sin(alpha)*cos(beta) + cos(alpha)*sin(beta)}}} ?

Apply it one time, and you will get

{{{sin(2alpha)}}} = {{{2*sin(alpha)*cos(alpha)}}},

{{{cos(2alpha)}}} = {{{cos^2(alpha) - sin^2(alpha)}}} = {{{1 - sin^2(alpha) - sin^2(alpha)}}} = {{{1 - 2*sin^2(alpha)}}}.

Apply it one more time, and you will get

{{{sin(3alpha)}}} = {{{sin(alpha)*cos(2alpha) + cos(alpha)*sin(2alpha)}}} = 
                  
          = {{{sin(alpha)*(1 - 2*sin^2(alpha)) + cos(alpha)*2*sin(alpha)*cos(alpha)}}} = 

          = {{{sin(alpha) - 2*sin^3(alpha) + 2*sin(alpha)*cos^2(alpha)}}} = 

          = {{{sin(alpha) - 2*sin^3(alpha) + 2*sin(alpha)*(1-sin^2(alpha))}}} = 

          = {{{3*sin(alpha) - 4*sin^3(alpha)}}}.
</pre>