Question

求解第一二题

Answer 1

1. f（x）=0.5sin2x·sinφ+2·0.5·（cos^2）x·cosφ-0.5cosφ
=0.5·[sin2x·sinφ+（2·（cos^2）x-1）cosφ]
=0.5·（sin2x·sinφ+cos2x·cosφ）
=0.5·cos（2x-φ）

2. f（x）=cos（π/3+x）·cos（π/3-x）-0.5·sin2x+0.25
=[cos（π/3）·cosx-sin（π/3）·sinx]·[cos（π/3）·cosx+sin（π/3）·sinx]-0.5·sin2x+0.25
=（0.5·cosx-√3/2·sinx）·（0.5·cosx+√3/2·sinx）-0.5·sin2x+0.25
=0.25（cos^2）x-0.75（sin^2）x-0.5·sin2x+0.25（cos^2）x+0.25（sin^2）x
=0.5（cos^2）x-0.5（sin^2）x-0.5·sin2x
=0.5cos2x-0.5sin2x
=0.5[sin（π/2+2x）+sin2x]
=0.5[sin（π/4+2x）·cos（π/4）]
=√2·sin（π/4+2x）

Answer 2

题不会，建议你用学霸君客户端

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