Question

2道简单高一数学题

1.已知sin(π/6-x)=1/4,sin(π/6+2x)=?
2.tan70cos10(1-√3tan20)=?
求简略的解答过程。

Answer 1

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

Answer 2

1、sin(π/6-x)=1/4，cos(π/6-x)=√15/4 或者 -√15/4 (因为 x=π/6-arcsin1/4 或者π/6-arcsin1/4+π/2)
∴ 2sin(π/6-x)*cos(π/6-x)= sin(π/3-2x)=cos[π/2-(π/3-2x)]=cos(π/6+2x)
即cos(π/6+2x)=√15/8 或者 -√15/8
当cos(π/6+2x)=√15/8，π/6+2x在第三象限，sin(π/6+2x)= -7/8
当cos(π/6+2x)= -√15/8，π/6+2x在第二象限，sin(π/6+2x)= 7/8

2、1-√3tan20=(cos20°-√3sin20°)/cos20°=2sin(30°-20°)/cos20°=2sin10°/cos20°
tan70°cos10°(1-√3tan20°)=tan70° *(2sin10°cos10°)/cos20°=cot70°tan20°=1