化简Sin(x/2)cos(x/2)+cos²(x/2)-2为Asin(λx+φ)+β
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化简Sin(x/2)cos(x/2)+cos²(x/2)-2为Asin(λx+φ)+β
化简Sin(x/2)cos(x/2)+cos²(x/2)-2为Asin(λx+φ)+β
化简Sin(x/2)cos(x/2)+cos²(x/2)-2为Asin(λx+φ)+β
原式= sin(x/2)cos(x/2)
= 1/2sinx+1/2cosx+1/2-2
= 1/2(sinx+cosx)-3/2
= (√2/2)sin(x+π/4)-3/2
其中,当形如asin(nx)+bcos(nx)时(注意,必需sin在前,cos在后),可化为√(a²+b²)sin(nx+β)
β前的正负由tan(b/a)的正负决定(注意,a、b的符号也要考虑)
β的值等于tan(b/a)所对应角的弧度,即用含π的数表示
如这道题,sinx+cosx= √(1²+1²)sin(x+π/4) (tan=1时对应角为π/4)
原式=1/2*sinx+(1+cosx)/2-2
=(1/2)(sinx+cosx)-3/2
=(√2/2)(√2/2*sinx+√2/2cosx)-3/2
=(√2/2)(sinxcosπ/4+cosxsinπ/4)-3/2
=(√2/2)sin(x+π/4)-3/2
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