三角函数求证sin1度sin2度sin3度……sin89度=(1/4)^45x6根10

Question

三角函数
求证sin1度sin2度sin3度……sin89度=(1/4)^45x6根10

Answer 1

利用三倍角公式
sin3a=4sinasin(60°+a)sin(60°-a)
证明如下：
sin3a=3sina-4sin^3a
=4sina(3/4-sin^2a)
=4sina[(√3/2)^2-sin^2a]
=4sina(sin^260°-sin^2a)
=4sina(sin60°+sina)(sin60°-sina)
=4sina*2sin[(60+a)/2]cos[(60°-a)/2]*2sin[(60°-a)/2]cos[(60°+a)/2]
=4sinasin(60°+a)sin(60°-a)
∴sin1度sin2度sin3度……sin89度
将1°,61°,59°一组,2°,62°,58°一组,.,29°,89°,31°一组,还剩30°和60°
=(1/4)^29* sin3°sin6°*sin9°*.*sin87°*(1/2)*(√3/2)
=√3*(1/4)^30* sin3°sin6°*sin9°*.*sin87°
将3°,63°,57°一组,6°,66°,54°一组,.,27°,87°,33°一组（共9组)还剩30°和60°
=√3*(1/4)^30*(1/4)^9 * sin9°sin18°*sin27°*.*sin81°*(1/2)*(√3/2)
=3*(1/4)^40* sin9°sin18°*sin27°*.*sin81°
=3*(1/4)^40 *(sin9°*cos9°)*(sin18°cos18°)*(sin27°*cos27°)*(sin36°*cos36°)*sin45°
=3*(√2/2)*(1/4)^42 *sin18°sin36°sin54°sin72°
=3*(√2/2)*(1/4)^42*sin18°cos18°*sin36°cos36°
=3*(√2/2)*(1/4)^43*sin36°*sin72°
=(3√2/2)*(1/4)^43* sin36°*cos18° (***)
∵ (sin36°cos18°)²
=(1/4)[(1-cos72°)(1+cos36°)]
=(1/4)*(1-cos72°+cos36°-cos36°cos72°)
=(1/4)(1+cos36°cos72°)
=(1/4)(1+sin36°cos36°cos72°/sin36°)
=(1/4)[1+(1/4)sin144°/sin36°]
=(1/4)*(5/4)
=5/16
∴ sin36°cos18°=√5/4
∴ sin1度sin2度sin3度……sin89度
=(3√2/2)*(1/4)^43* sin36°*cos18°
=(3√2/2)*(1/4)^43*(√5/4)
=(3√10/2)*(1/4)^44
=6√10 *(1/4)^45
∴ 原等式成立.