求证:cos(n-1)AcosA-cosnA = sin(n-1)AsinA .

问题描述:

求证:cos(n-1)AcosA-cosnA = sin(n-1)AsinA .
还有 已知sin( A + 24°) = sin( A + 66°),求tanA的值。

cosna=cos[(n-1)a+a]=cos(n-1)acosa-sin(n-1)acsinacos(n-1)acosa-cosna=cos(n-1)acosa-{cos(n-1)acosa-sin(n-1)acsina}=sin(n-1)acsina sin(A+66)=sin(A+24) sin(A+66)-sin(A+24)=0 2cos(A+45)sin21=0,(sin21≠0)...