Question

соsa+cos2a+cos6a+cos7a=

Answer 1

cosα+cos2α+cos6α+cos7α=(cosα+cos7α)+(cos2α+cos6α)=
=2·cos(α+7α)/2·cos((α-7α)/2+2·cos(2α+6α)/2·cos(2α-6α)/2=
=2cos4α·cos(-3α)+2·cos4α·cos(-2α)=
=2·cos4α·(cos3α+cos2α)=2cos4α·2·cos5α/2·cosα/2=
=4cos4α·cos5α/2·cosα/2;