作业帮形如SIN^2(X) + SIN2X + COS^2X 化简...y=sin^2 x+sin 2x 3cos^2 x6sin^2x+sin x cos x-2cos^ x=0求sin(2x+60)的值送分了。怎么没人做呢
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形如SIN^2(X) + SIN2X + COS^2X 化简...y=sin^2 x+sin 2x 3cos^2 x6sin^2x+sin x cos x-2cos^ x=0求sin(2x+60)的值送分了。怎么没人做呢
形如SIN^2(X) + SIN2X + COS^2X 化简...
y=sin^2 x+sin 2x 3cos^2 x
6sin^2x+sin x cos x-2cos^ x=0
求sin(2x+60)的值
送分了。怎么没人做呢
形如SIN^2(X) + SIN2X + COS^2X 化简...y=sin^2 x+sin 2x 3cos^2 x6sin^2x+sin x cos x-2cos^ x=0求sin(2x+60)的值送分了。怎么没人做呢
=(sinx)^2+2sinxcosx+(cosx)^2
=(sinx+cosx)^2
=2*[sin(x+π/4)]^2
题目没说太清楚
y=sin^2 x+sin 2x 3cos^2 x
y=(1-cos2x)/2+sin2x * 1/2(1+cos2x)
=1/2-1/2cos2x+1/2sin2x+1/4sin4x
=√2/2*sin(2x+π/4)+1/4sin(4x)+1/2
至于第二题，我感觉有毛病，按题目解答
6sin^2x+sin x cos x-2cos^ 2x=0
...
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y=sin^2 x+sin 2x 3cos^2 x
y=(1-cos2x)/2+sin2x * 1/2(1+cos2x)
=1/2-1/2cos2x+1/2sin2x+1/4sin4x
=√2/2*sin(2x+π/4)+1/4sin(4x)+1/2
至于第二题，我感觉有毛病，按题目解答
6sin^2x+sin x cos x-2cos^ 2x=0
6tan^2x+tanx-2=0
tanx+1/12=+1/12或-1/12
tanx=0,tanx=-1/12
tanx=0,x=90*k,
sin(2x+60)=sin(180K+60)=√3/2
当tanx=-1/12,你分别求出来sinx,cosx就可以了
我个人感觉这个题目有问题，按说就是6sin^2x+sin x cos x-2cos^ 2x=0
化简后可以与sin(2x+60)拆开的式子联系上，就可以解题了，希望能帮到你
收起
SIN^2(X) + SIN2X + COS^2X =SIN^2(X) + 2SINX COSX+ COS^2X =(SINX+COSX)^2=2SIN^2(X+π/4)