Prove that:
[tex]cos(2\alpha )-cos(8\alpha )=sin(5\alpha )sin(3\alpha )[/tex]
Remember that:
[tex]cos(A+B)=cos(A)cos(B)-sin(A)sin(B)\\cos(A-B)=cos(A)cos(B)+sin(A)sin(B)\\\\\\cos(A+B)-cos(A-B)=-2sin(A)sin(B)[/tex]
We wish to remove this add (A+B).
[tex]\left \{ {{A+B=p} \atop {A-B=q}} \right. \\\\2A=p+q[/tex]
[tex]A=\frac{p+q}{2} \\\\B=\frac{p-q}{2}[/tex]
Then we de identity:
[tex]cos(p)-cos(q)=-2sin(\frac{p+q}{2}) sin(\frac{p-q}{2})[/tex]
Let p=2α and q=8α
[tex]cos(2\alpha )-cos(8\alpha )=-2sin(\frac{2\alpha +8\alpha }{2})sin(\frac{2\alpha -8\alpha }{2}) \\\\cos(2\alpha )-cos(8\alpha )=-2sin(5\alpha )sin(-3\alpha )\\\\*sin(-\alpha )=-sin(\alpha )\\\\cos(2\alpha )-cos(8\alpha )=2sin(5\alpha )sin(3\alpha )[/tex]
