f(x)=1/2 sin2xsinφ+cos^2xcosφ-1/2sin(π/2+φ),其中φ = π/3 .
问题描述：
f(x)=1/2 sin2xsinφ+cos^2xcosφ-1/2sin(π/2+φ),其中φ = π/3 .
横坐标缩短为原来的1/2,纵坐标不变,得到g(x),g(x)=1/2cos(4x-π/3)
g(x)=1/2cos(4x-π/3)怎么算出来的?
答
f(x)=1/2 sin2xsinφ+cos^2xcosφ-1/2sin(π/2+φ),
=1/2 sin2xsinφ+1/2(1+cos2x)cosφ-1/2sin(π/2+φ),
=1/2 sin2xsinφ+1/2cos2xcosφ-1/2sin(π/2+φ)+1/2
=1/2cos(2x-φ)-1/2sin(π/2+φ)+1/2 (φ = π/3 )
=1/2cos(2x-π/3)-1/2sin5π/6)+1/2
=1/2cos(2x-π/3)+1/2
横坐标缩短为原来的1/2,周期为原来一半
原来T=2π/2=π,现在T=π/2
所以f(x)=1/2cos(4x-π/3)+1/2 (T=2π/4=π/2),明白吗?