sin(x-3兀/4)cos(x-兀/4)=-1/4,求cos4x
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sin(x-3兀/4)cos(x-兀/4)=-1/4,求cos4x
sin(x-3兀/4)cos(x-兀/4)=-1/4,求cos4x
sin(x-3兀/4)cos(x-兀/4)=-1/4,求cos4x
sin(x-3π/4) cos(x - π/4) = - 1/4,求cos4x
将要用到:2sinAcosB = sin(A+B) + sin(A-B)
和 cos2α = cos²α - sin²α = 1 - 2sin²α
sin(x-3π/4) cos(x - π/4) = 1/2 [sin(2x -π) + sin(-π/2)] = -1/2 [sin2x + 1] = -1/4
===> sin2x = -1/2
cos4x = 1 - 2 sin²2x = 1 - 2 *(1/4) = 1 - 1/2 = 1/2
这样做最简洁的,当然楼上几位的方法也值得参考!
化简出来 (sinx+cosx)^2=1/2 1+2sinxcosx=1/2 sin2x=-1/2 cos4x=1-2(sin2x)^2=1/2
原式=[sinx*cos(3π/4) - cosx*sin(3π/4)]*[cosx*cos(π/4) + sinx*sin(π/4)]
=[(-√2/2)sinx - (√2/2)cosx]*[(√2/2)cosx + (√2/2)sinx]
=-[(√2/2)sinx + (√2/2)cosx]*[(√2/2)cosx + (√2/2)sinx]
=-(√2/2)^2*...
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原式=[sinx*cos(3π/4) - cosx*sin(3π/4)]*[cosx*cos(π/4) + sinx*sin(π/4)]
=[(-√2/2)sinx - (√2/2)cosx]*[(√2/2)cosx + (√2/2)sinx]
=-[(√2/2)sinx + (√2/2)cosx]*[(√2/2)cosx + (√2/2)sinx]
=-(√2/2)^2*(cosx + sinx)^2
=-(1/2)[(cosx)^2 + (sinx)^2 + 2ainxcosx]
=-(1/2)(1 + sin2x) = -1/4
则:1 + sin2x = 1/2
sin2x=-1/2
cos4x =1 - 2*(sin2x)^2 =1 - 2*(-1/2)^2 =1 - (1/2) =1/2
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sin(x-3兀/4)cos(x-兀/4)=-1/4
sin(x+兀/4-兀)cos(x-兀/4)=-1/4
-sin(x+兀/4)cos(x-兀/4)=-1/4
sin(x+兀/4)cos(x-兀/4)=1/4
(sinxcos兀/4+sin兀/4cosx)(cosxcos兀/4+sinxsin兀/4)=1/4
√2/2(sinx+cosx)*√2/2(s...
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sin(x-3兀/4)cos(x-兀/4)=-1/4
sin(x+兀/4-兀)cos(x-兀/4)=-1/4
-sin(x+兀/4)cos(x-兀/4)=-1/4
sin(x+兀/4)cos(x-兀/4)=1/4
(sinxcos兀/4+sin兀/4cosx)(cosxcos兀/4+sinxsin兀/4)=1/4
√2/2(sinx+cosx)*√2/2(sinx+cosx)=1/4
1/2(sinx+cosx)^2=1/4
(sinx+cosx)^2=1/2
1+2sinxcosx=1/2
sin2x=-1/2 (2sinxcosx=sin2x)
(sin2x)^2=1/4
所以cos4x=1-2(sin2x)^2=1-2*(1/4)=1/2
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sin(x-3兀/4)cos(x-兀/4)=(1/2){sin[(x-3兀/4)+(x-兀/4)]+sin[(x-3兀/4)-(x-兀/4)]}=-1/4
sin(2x-兀)+sin(-兀/2)=-1/2,-sin(兀-2x)-1=-1/2,sin2x=-1/2,cos4x=1-2(sin2x)^2=1-2*(1/4)=1/2