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已知A,B,C是△ABC的三个内角,F(x)=Cos2x+5/2sinAsin...

∵A+π属于(π/2,3π2) 又sin(A+π/4)=7根号2/10 ∴cos(A+π/4)=-根号2/10 sinA=sin[(A+π/4)-π/4]=4/5 f(x)=cos2x+5/2*4/5*sinx=1-2sin²x+2sinx=-2(sinx-1/2)²+3/2 当sinx=1/2时f(x)max=3/2 当sinx=-1时f(x)min=-3

∵ f(x) = 2x²+4xsinA+3cosA 有且只有一个零点 ∴ △ = (4sinA)²-4*2*3cosA=0 ∴ 2sin²A-3cosA=0 ∴ 2-2cos²A-3cosA=0 2cos²A+3cosA-2=0 (cosA+2)(2cosA-1)=0 2cosA-1=0 cosA=1/2 A=60° 第二问: S=3√3 1/2bcsinA=3√3 bc=6√...

cosx^2-sinx^2+2sinxcox=(cosx-sinx)^2

你好: 解:f(X)=cos(2x+π/3)+sin^2x =cos2xcos60-sin2xsinπ/3 =1/2cos2x-√3/2sin2x+1/2-cos2x/2 =-√3/2sin2x+1/2 所以当sin2x=-1时,f(x)取最大值:√3/2+1/2 T=2π/W=2π/2=π f(c/2)=-√3/2sinc+1/2=-1/4 推出:sinc=√3/2 C=60度 因为:cosB=-...

(1)∵sinC=2sinA,由正弦定理可得:c=2a又∵b=3a,∴cosB=a2+c2?b22ac=4a2+a2?(3a)22?2a?a=12∴B=π3(2)S△ABC=12?ac?sinB=12×a×2a?sinπ3=23∴a=2∴f(x)=2sin2x+cos(2x?B)?a=1?cos2x+cos(2x?π3)?2=?cos2x+12cos2x+32sin2x?1=32sin2x?12co...

亲,你传个手写的吧,这个看不太明白

cos(A-C)-cos(A+C), cosAcosC+sinAsinC-(cosAcosC-sinAsinC) =2sinAsinC=2sinB 所以sinAsinC=sinB f(C)=2sin(2C+π/3)=√3 所以sin(2C+π/3)=√3/2 即2C+π/3=60°或120° 当2C+π/3=60°,C=0 不符舍去 所以2C+π/3=120°,C=30° sinAsinC=sinB=sin(A+C)...

解:f(x)=a*b=sinx(6sinx+cosx)+cosx(7sinx-2cosx)=8sinxcosx+6(sinx)^2-2(cosx)^2 =4sin2x+6-8(cosx)^2=4sin2x+6-4*(cos2x+1)=4sin2x-4cos2x+2 =4√2(sin2xcosπ/4-cos2xsinπ/4)+2=4√2sin(2x-π/4)+2≤4√2+2 f(A)=6=4√2sin(2A-π/4)+2 得sin(2A-π/4)...

(1)∵在△ABC中,sinA=513,cosB=35,∴cosA=1213,sinB=45,∴cosC=cos[π-(A+B)]=-cos(A+B)=-cosAcosB+sinAsinB=-1213×35+513×45=?1665.∴cosC=?1665.(2)∵sin2x+2sin2x1?tanx=2sinxcosx+2sin2x1?sinxcosx=2sinxcosx(cosx+sinx)cosx?sinx=2...

(1)f(X)=cos(2x+pai/3)+sin^2x =cos2xcos60-sin2xsinpai/3 =1/2cos2x-根号3/2sin2x+1/2-cos2x/2 =-根号3/2sin2x+1/2 所以当sin2x=-1时,f(x)取最大值:根号3/2+1/2 T=2pai/W=2pai/2=pai (2)f(c/2)=-根号3/2sinc+1/2=-1/4 推出:sinc=根...

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