2cos x (sin x -cos x)+1
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2cosx(sinx-cosx)+12cosx(sinx-cosx)+12cosx(sinx-cosx)+12cosx(sinx-cosx)+1=2sinxcosx-2cosx^2+1=sin2x+1
2cos x (sin x -cos x)+1
2cos x (sin x -cos x)+1
2cos x (sin x -cos x)+1
2cosx (sinx -cosx)+1 =2sinxcosx-2cosx^2+1
=sin2x+1-2cosx^2
=sin2x-cos2x
=√2sin(2x-π/4)
直接打开括号,1-2cx=-c2x 2cs=s2x s2x-c2x=√2s(2x-45) 如果你能告诉我砸打那些符号我会很感谢首尾
补充的。第一题确实有问题。 7。由图形知,只需保证a-1
原式=sin2x+1-2cosx^2
=sin2x-cos2x
=(根号2)sin(2x-π/4)
原式=2cosxsinx-2cosxcosx+1=sin2x+(1-2cosxcosx)=sin2x-cos2x
2cos x (sin x -cos x)+1
y=2cos x (sin x+cos x)
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