函数f(x)=sinx cosx的最小值是(求过程)
2个回答
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解:
f(x) = [sin²x+cos²x + 2sincosx - (sin²x+cos²x)] / 2
=[(sinx+cox)² - 1] / 2
=(1/2)(sinx+cox)² - (1/2)
因为:sinx+cox = √2 sin(x+ π/4)
f(x) = sin²(x+ π/4) - (1/2)
根据三角函数性质:
0≤sin²(x+ π/4)≤1
因此:
f(x) = sin²(x+ π/4) - (1/2)的最小值为:1/2
f(x) = [sin²x+cos²x + 2sincosx - (sin²x+cos²x)] / 2
=[(sinx+cox)² - 1] / 2
=(1/2)(sinx+cox)² - (1/2)
因为:sinx+cox = √2 sin(x+ π/4)
f(x) = sin²(x+ π/4) - (1/2)
根据三角函数性质:
0≤sin²(x+ π/4)≤1
因此:
f(x) = sin²(x+ π/4) - (1/2)的最小值为:1/2
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