Question

sin(π-α)+sin(2π-α)+sin(3π-α)+sin(4π-α)=______________.

Answer 1

解析:sin(π-α)+sin(2π-α)+sin(3π-α)+sin(4π-α)=sinα-sinα+sin(π-α)+sin(-α)=sinα-sinα+sinα-sinα=0.

答案:0