cos(x+1)½-cosx½→-2sin{[(x+1)½+x½]/2}*sin{[(x+1)½-x½]}如何转化
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![cos(x+1)½-cosx½→-2sin{[(x+1)½+x½]/2}*sin{[(x+1)½-x½]}如何转化](/uploads/image/z/12543816-48-6.jpg?t=cos%28x%2B1%29%26%23189%3B-cosx%26%23189%3B%E2%86%92-2sin%7B%5B%28x%2B1%29%26%23189%3B%2Bx%26%23189%3B%5D%2F2%7D%2Asin%7B%5B%28x%2B1%29%26%23189%3B-x%26%23189%3B%5D%7D%E5%A6%82%E4%BD%95%E8%BD%AC%E5%8C%96)
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cos(x+1)½-cosx½→-2sin{[(x+1)½+x½]/2}*sin{[(x+1)½-x½]}如何转化
cos(x+1)½-cosx½→-2sin{[(x+1)½+x½]/2}*sin{[(x+1)½-x½]}如何转化
cos(x+1)½-cosx½→-2sin{[(x+1)½+x½]/2}*sin{[(x+1)½-x½]}如何转化
a=(x+1)½, b=x½
cos(x+1)½=cosa=1-2sin^2(a/2)
cosx½=cosb=1-2sin^2(b/2)
cos(x+1)½-cosx½=1-2sin^2(a/2)-[1-2sin^2(b/2)]
=2sin^2(b/2)-2sin^2(a/2)
=2[sin(b/2)+sin(a/2)][sin(b/2)-sin(a/2)]
=2{sin(x½/2)+sin[(x+1)½/2]}*[sin(x½/2)-sin[(x+1)½/2]}
=-2{sin(x½/2)+sin[(x+1)½/2]}*{[sin[(x+1)½/2]-sin(x½/2)}