sin3＜sin（-5）＜sin5,

对,sin5 > sin3 > 0
sin(- 5) = - sin5 < 0
所以sin(- 5) < sin3 < sin5

用a
cos3a=cos(2a+a)
=cos2acosa-sin2asina
=(2cos²a-1)cosa-2sinacosa*sina
=2cos³a-cosa-2(1-cos²a)cosa
=4cos³a-3cosa
所以分子=4cosa-4cos³a
=4cosa(1-cos²a)
=4cosasin²a
同理
sin3a=3sina-4sin³a
分母=2sina-4sin³a
=2sina(1-2sin²a)
原式=2sinacosa/(1-2sin²a)
=sin2a/coa2a
=tan2a

有一个结论
sin(π/n)*sin(2π/n)*sin(3π/n)*…*sin[(n-1)π/n] = n/ 2^(n-1)
于是
sin(π/7)*sin(2π/7)*sin(3π/7)*sin(4π/7)*sin(5π/7)*sin(6π/7) = 7/64
sin(π/7)*sin(2π/7)*sin(3π/7)*sin(3π/7)*sin(2π/7)*sin(π/7) = 7/64
sin(π/7)*sin(2π/7)*sin(3π/7) = √7 /8

答案:0
解法：
因为sin5兀=sin兀=0
所以任何数与(sin5兀)乘积为零
所以sin兀/14*sin3兀/14*sin5兀=0
常规解法用积化和差法
sin兀/14*sin3兀/14*sin5兀
=-1/2*(cos4兀/14-cos2/14兀)*sin兀
=0

sin3a -sina = sin(a+2a) + sin(-a) = sin(a+2a) + sin(a-2a)
= 2sinacos2a
cos2a = 1-2sin²a = 1- 2*1/9 = 7/9
sin3a -sina = 2 * 1/3 * 7/9 = 14/27
sina = 14/27 + 1/3 = 14/27 + 9/27 = 23/27

sin3π/4=√2/2
sin11π/4=√2/2
sin7π/4=-√2/2
sin-1π/2=-1
sin-1π/4=-√2/2

（3sina+sin3a)/(3cosa+cos3a)
=(3sina+3sina-4sin^3a)/(3cosa+4cos^3a-3cosa)
=(6sina-4sin^3a)/(4cos^3a)
=3/2tana*1/cos^2a-tan^3a
=3/2a*(1+tan^2a)-a^3
=3/2a+3/2a^3-a^3
=3/2a+a^3/2

π>3π/5>4π/7>3.5π/7,他们都在第二象限,角度大的数值小,故
sin3π/5小
sin4π/7
大

LOG4 SIN3/4π+LOG3 tan(-5/6π)
=-1/4-1/2
=-3/4