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已知函数F(x)=sin(3x+π/4) F(A/3)=4/5Cos(A+π/4)Co...

接下来应该需要用到积化和差之类的公式,我对这类公式不熟悉,只能做到这一步了。望采纳。

详解过程!你参考看看~

ok

cos(a+π/4)=3/5=cosa×cos(π/4)-sina×sin(π/4)=√2/2×(cosa-sina)=3/5 ∴cosa-sina=(3√2)/5…………① ①²=cos²a+sin²a-2sinacosa=1-2sinacosa=18/25 ∴ 2sinacosa=7/25 sin²a+cos²a+2sinacosa=1+7/25=(sina+cosa)²=32/25 ∵...

(1) f(x)=4cosxsin(x- π/3)+a =2[sin(x+x-π/3)-sin(x-x+π/3)]+a =2sin(2x -π/3) +a-√3 最小正周期T=2π/2=π (2) sin(2x -π/3) =1时,f(x)取得最大值 f(x)max=2+a-√3=2 a=√3 f(x)=2sin(2x -π/3) A

①.cos2a=cos((a+b)+(a-b))=cos(a+b)*cos(a-b)-sin(a+b)*sin(a-b) 又因为3/2π小于a+b小于2π,即在第四象限,π/2小于a-b小于π,即在第二象限 所以sin(a+b)0,sin(a-b)0 值=(4/5)*(-4/5)-(3/5)*(-3/5)=-7/25 ②.第二题b是π/2bπ吧? tana+tanb=5/6 tan...

f(x)=cox(2x-π/3)+2sin(x-π/4)sin(x+π/4) =cox(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)] =cox(2x-π/3)+2sin(x-π/4)cos(x-π/4) =cox(2x-π/3)+sin(2x-π/2) =cox2x*cosπ/3+sin2x*sinπ/3-cos2x =1/2*cos2x+sin2x*sinπ/3-cos2x =sin2x*sinπ/3-1/2cos2x =...

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

cos(π/4-a)=4/5,cosπ/4cosa+sinπ/4sina=4/5,cosa+sina=4√2/5,1-sin²a=32/25-8√2sina/5+sin²a,2sin²a-8√2sina/5+7/25=0,sina=(8√2/5±6√2/5)/4,sina=7√2/10或sina=√2/10,∵π/4

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