三角函数数学题一道,关于倍角半角求值：log2 cos(1/9π) + log2 cos(2/9π) + log2 cos(4/9π)注意2是下标答案是-3,就是不知过程,敬请提示或指教,
问题描述：
三角函数数学题一道,关于倍角半角
求值：log2 cos(1/9π) + log2 cos(2/9π) + log2 cos(4/9π)
注意2是下标
答案是-3,就是不知过程,敬请提示或指教,
答
作和后，真数部分乘以sin(π/9)再算:
log2 cosπ/9+ log2 cos2π/9+ log2 cos4π/9
=log2 [(cosπ/9)*(cos2π/9)*(cos4π/9)]
=log2 [(sinπ/9)*(cosπ/9)*(cos2π/9)*(cos4π/9)/(sinπ/9)]
=log2 [(1/8)*sin(8π/9)/(sinπ/9)]
=log2 (1/8)
=-3
肯定对的，放心
答
等价于计算COS(1/9TT)*COS(2/9TT)*COS(4/9TT) 将它乘上2SIN（1/9TT） 后来在 自己区做做 可以做的
答
作和后,真数部分乘以sin(π/9)再算:
log2 cosπ/9+ log2 cos2π/9+ log2 cos4π/9
=log2 [(cosπ/9)*(cos2π/9)*(cos4π/9)]
=log2 [(sinπ/9)*(cosπ/9)*(cos2π/9)*(cos4π/9)/(sinπ/9)]
=log2 [(1/8)*sin(8π/9)/(sinπ/9)]
=log2 (1/8)
=-3