我才能做下一部求周期和单调区间.
1个回答
简单,
y = 1/2 + sin²x + sinxcosx
= 1 - 1/2(1-2sin²x) + sinxcosx·······································式1
根据倍角公式 sin2a = 2sina•cosa 半角公式 cos2a = cos²a-sin²a = 1-2sin²a
则式1可变化为：
y =1 - 1/2cos2x + 1/2sin2x
= 1- 1/2( cos2x - sin2x )················································式2
化简其中的cos2x - sin2x
根据诱导公式：sinα = -cos(π/2+α) ,可知：sin2x = -cos(π/2+2x)
则式2变化为：
y = 1- 1/2[ cos(π/2+2x) + cos2x ]···································式3
根据和差化积公式：cosθ+cosφ = 2 cos[(θ+φ)/2]cos[(θ-φ)/2]
化简式3:,可得
y = 1 - 1/2 cos(2x+π/4)cos(π/4)
= 1 - √2/4 cos(2x+π/4)···············································注意√2表示根号2
化简完毕
