1.已知sin(π/6-x)=1/4,sin(π/6+2x)=?

题目

1.已知sin(π/6-x)=1/4,sin(π/6+2x)=?
2.tan70cos10(1-√3tan20)=?

答案

sin(PI/6 + 2x) = cos (PI/2 - PI/6 - 2x) = cos(PI/3 - 2x)
= cos( 2* (PI/6 - x) ) = 1 - 2 * sin(PI/6 - x) ^ 2 = 1 - 2 * (1/4)^2 = 7/8
tan70 * cos10 * (1-√3tan20)
= tan70 * cos10 - √3 cos10
= cos20/sin20 * cos10 - √3 cos10
= cos20/(2sin10) - √3 cos10
= (cos20 - 2√3sin10cos10) / 2sin10
= 1/2 * (cos20 - √3sin20) / sin10
= (1/2 * cos20 - 1/2 * √3 sin20) / sin10
= (sin30*cos20 - cos30*sin20) / sin10
= sin(30-20) / sin10
= 1