sin^220°+cos^250°+sin30°sin70°
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/26 11:07:55
sin^220°+cos^250°+sin30°sin70°sin^220°+cos^250°+sin30°sin70°sin^220°+cos^250°+sin30°sin70°(sin20)^2+
sin^220°+cos^250°+sin30°sin70°
sin^220°+cos^250°+sin30°sin70°
sin^220°+cos^250°+sin30°sin70°
(sin20)^2+(cos50)^2+sin30*sin70
=(sin20)^2+(cos50)^2+sin(50-20)*sin(50+20)
=(sin20)^2+(cos50)^2+(sin50cos20-cos50sin20)*(sin50cos20+cos50sin20)
=(sin20)^2+(cos50)^2+(sin50)^2*(cos20)^2-(cos50)^2*(sin20)^2
=(sin20)^2*[1-(cos50)^2]+(cos50)^2+(sin50)^2*(cos20)^2
=(sin20)^2*(sin50)^2+(cos50)^2+(sin50)^2*(cos20)^2
=(sin50)^2*[(sin20)^2+(cos20)^2]+(cos50)^2
=(sin50)^2+(cos50)^2
=1
约等于：0.117+0.117+0.4698≈0.7038
sin^220°+cos^250°+sin20°cos50°
sin^220°+cos^250°+sin30°sin70°
化简3/(sin^220°)-1/(cos^220°)+64sin^220°
sin(10°)+sin(20°)cos(50°)求值
cos^15°-sin^15°=
sin(α+30°)-sin(α-30°)/cosα
sin^2 30°+cos()=1
计算：sin^67.5°-cos^67.5度
sin²15° 把sin变成带cos的
观察下列等式：①sin²10°＋cos²40＋sin10°cos40°＝3／4；②sin²6°＋cos²36°＋si②sin²6°＋cos²36°＋sin6°cos36°＝3／4。由上面两式的结构规律，你是否能提出一个猜想？并证明
sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
Cos(180°+α)*sin（α+360°)/sin(-α-180°)*cos(-180°-α)
sin(33°-x)cos(27°+x)+cos(33°-x)sin(27°+x)
sin(-120°)cos(1290°)+cos(-1020°)sin(-1050°)=
sin(70° a)cos(10° a)-cos(70° a)sin(170°-a)
sin(a+45°)sin(a+15°)+cos(a+45°)cos(a+15°)
cos(α+30°)cosα+sin(α+30°)sinα等于?
sin(α+30°)cosα+cos(α+30°)sinα=