解
sinθ-2cosθ=0
∴sinθ=2cosθ
∴tanθ=sinθ/cosθ=2
∴sin²θ+sinθcosθ-2cos²θ
=(sin²θ+sinθcosθ-2cos²θ)/(sin²θ+cos²θ)————(除以cos²θ+sin²θ=1,值不变)
=(tan²θ+tanθ-2)/(tan²θ+1)————(分子分母除cos²θ)
=(4+2-2)/(4+1)
=4/5
若sinθ-2cosθ=0,求:sin^2θ+sinθcosθ-2cos^θ的值
若sinθ-2cosθ=0,求:sin^2θ+sinθcosθ-2cos^θ的值
求详解,谢谢啊
求详解,谢谢啊
数学人气:805 ℃时间:2020-04-08 02:52:31
优质解答
我来回答
类似推荐
猜你喜欢
- 1看图写话,故事是小明在捉一只老鼠
- 24x-b=5的解为x=2,则直线y=4x-b的图像一定经过点( )
- 3若函数f(x)=kx2+(k-1)x+2是偶函数,则f(x)的递减区间是_.
- 4Our favorite food 为题,谢一篇短文.要求语句通顺,字数约50个单词
- 5英语笑话短文
- 6Excause me,when__you__the man in bule? Lst year in Nanjing.And I ___ him for about one year
- 7The garden _______while the Greens were away from home.
- 8数学的√数怎么算?
- 9What are you going to do tomorrow?_________空上选哪项?为什么?
- 10怎样用六根小棒围成四个三角形(一样长)?