Question

докажите тождество
4sin^6a+4cos^6a-1=3cos2a
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Answer 1

$4Sin^{6}\alpha+4Cos^{6} \alpha -1=4(Sin^{6}\alpha +Cos^{6}\alpha)-1=4[(Sin^{2}\alpha)^{3} +(Cos^{2}\alpha)^{3}]-1=\\\\=4[\underbrace{(Sin^{2} \alpha+Cos^{2} \alpha)}_{1}( Sin^{^{4} }\alpha-Sin^{2}\alpha Cos^{2}\alpha+Cos^{4}\alpha)]-1=\\\\=4[(Sin^{2}\alpha+Cos^{2} \alpha)^{2} -3Sin^{2} \alpha Cos^{2}\alpha]-1=4(1-3Sin^{2} \alpha Cos^{2}\alpha)-1=\\\\=4-12Sin^{2} \alpha Cos^{2}\alpha-1=3-12Sin^{2} \alpha Cos^{2}\alpha=3(1-4Sin^{2} \alpha Cos^{2}\alpha)=$

$=3(1-Sin^{2} 2\alpha)=\boxed{3Cos^{2}2\alpha}$