Question

已知sinx=2cosx,则sin^2x+1=

Answer 1

解：sinx=2cosx∴tanx=2∴sin�0�5x+1=(2sin�0�5x+cos�0�5x)/(sin�0�5x+cos�0�5)分号两边同除以cos�0�5x=(2tan�0�5x+1)/(tan�0�5x+1)=(2×2�0�5+1)/(2�0�5+1)=9/5

Answer 2

∵sinx=2cosx ∴tanx=sinx/cosx=2sin�0�5x+1=(sin�0�5x+1)/1=(sin�0�5x+sin�0�5x+cos�0�5x)/(sin�0�5x+cos�0�5x)分式上下同除cos�0�5x得sin�0�5x+1=(2tan�0�5x+1)/(tan�0�5x+1)=9/5

Answer 3

tanx=2(sin^2 x+sin^2 x +cos^2 x)/(sin^2 x +cos^2 x)=同时除以COS^2 X得：(tan^2 x+tan^2 x+1)/（tan^2 x+1）带入tanx=2得原式等于（4+4+1）/（4+1）=9/5