Question

求log2cos3.14/9+log2cos6.28/9+log2cos12.56/9的求值过程。(注:运用高一的数学知识求解)

数学题，如上.

Answer 1

log2cos3.14/9+log2cos6.28/9+log2cos12.56/9
=log2(cos3.14/9*cos6.28/9*cos12.56/9)
=log2(cos3.14/9*2*sin3.14/9cos6.28/9*cos12.56/9/2*sin3.14/9)
=log2(sin6.28/9*cos6.28/9*cos12.56/9/2*sin3.14/9)
=log2(2*sin6.28/9*cos6.28/9*cos12.56/9/2*2*sin3.14/9)
=log2(sin12.56/9*cos12.56/9/2*2*sin3.14/9)
=log2(2*sin12.56/9*cos12.56/9/2*4*sin3.14/9)
=log2(sin25.12/9/8sin3.14/9)
=log2(sin3.14/9/8sin3.14/9)
=log2(1/8)
=-3

Answer 2

原式=log2[cos(pi/9)*cos(2pi/9)*cos(4pi/9)]
先只计算cos(pi/9)*cos(2pi/9)*cos(4pi/9)
在式子前面乘以sin(pi/9)的同时，在式子底下添上分母sin(pi/9)
然后分子是sin(pi/9)*cos(pi/9)*cos(2pi/9)*cos(4pi/9)
=1/2*sin(2pi/9)*cos(2pi/9)*cos(4pi/9)
=1/4*sin(4pi/9)*cos(4pi/9)
=1/8*sin(8pi/9)
分母是sin(pi/9)=sin(pi-8pi/9)=sin(8pi/9)
分子除分母得1/8
原式就变成了log2(1/8)=-log2(8)=-3

Answer 3

计算器最好