设tan（α+8/7π）=a 求证sin（22π/7+α）+3cos（α-20π/7）/ sin（20π/7-α）-cos（α+22π/7）=a+3/-a

Question

设tan（α+8/7π）=a 求证sin（22π/7+α）+3cos（α-20π/7）/ sin（20π/7-α）-cos（α+22π/7）=a+3/-a

Answer 1

tan(α+8π/7)=tan(π+α+π/7)=tan(α+π/7),即：tan(α+π/7)=a
sin（22π/7+α）+3cos（α-20π/7）/ sin（20π/7-α）-cos（α+22π/7）
=(sin(3π+π/7+α)+3cos(3π-2π/7-α))/(sin(3π-2π/7-α)-cos(3π+π/7+α))
=－(sin(π/7+α)+3cos(2π/7+α))/(sin(2π/7+α)+cos(π/7+α))
=－(sin(π/7+α)+3(cosπ/7cos(π/7+α)-sinπ/7sin(π/7+α)))/(sinπ/7cos(π/7+α)+cosπ/7sin(π/7+α)+cos(π/7+α))
=－(tan(π/7+α)+3(cosπ/7-sinπ/7tan(π/7+α)))/(sinπ/7+cosπ/7tan(π/7+α)+1)
=－(a+3(cosπ/7-asinπ/7))/(sinπ/7+acosπ/7+1)