sin²(π/4+x)怎样等于cos²(π/4-x)?
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![sin²(π/4+x)怎样等于cos²(π/4-x)?](/uploads/image/z/1746965-29-5.jpg?t=sin%26%23178%3B%EF%BC%88%CF%80%2F4%2Bx%29%E6%80%8E%E6%A0%B7%E7%AD%89%E4%BA%8Ecos%26%23178%3B%EF%BC%88%CF%80%2F4-x%29%3F)
sin²(π/4+x)怎样等于cos²(π/4-x)?
sin²(π/4+x)怎样等于cos²(π/4-x)?
sin²(π/4+x)怎样等于cos²(π/4-x)?
sin(π/4+x)=cos(π/2-(π/4+x)=cos(π/4-x)函数名奇变偶不变
再平方∴sin²(π/4+x)=cos²(π/4-x)
∵sin(π/4+x)
=cos[π/2-(π/4+x)]
=cos(π/4-x)
∴sin²(π/4+x)=cos²(π/4-x)