设sin2a=a,cos2a=b,0小于b小于π 4,给出tan(π 40
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因为:tana/2=1/2所以:tana=2tan(a/2)/[1-tan²(a/2)]=4/3则:(1+sin2a)/(1+sin2a+cos2a)=(sin²a+cos
(sin2a+cos2a-1)(sin2a-cos2a+1)=sin²2a-(cos2a-1)²=-2cos²2a+2cos2a=根号3sin4a,即-2cos²
sin2a=2sinacosa1=(sina)^2+(cosa)^22(sin2a+1)=2(sina+cosa)^21+sina2a+cos2a=1+2sinacosa+2(cosa)^2-1=2c
左边平方差=(sin²a+cos²a)(sin²a-cos²a)因为sin²a+cos²a=1所以左边=sin²a-cos&sup
2(sin2a+1)/1+sin2a+cos2a=2(sina+cosa)2/[2cosa(sina+cosa)]=(sina+cosa)/cosa=tana+1
(sin²a+sin2a)/(cos²a+cos2a)=(sin²a+2sinacosa)/(cos²a+cos²a-sin²a)=(sin
这个等式是不成立的.假设a=30,那么sin4a-cos4a=sin120-cos120=sin60+cos60=/2sin2a-cos2a=sin60-cos60=/2
sin2a+sina/cos2a+cosa+1=(2sinacosa+sina)/(2cos²a+cosa)=sina(2cosa+1)/[cosa(2cosa+1)]=sina/cosa=
证明:1-cos(2A)=2*[(sinA)^2]1+cos(2A)=2*[(cosA)^2]sin(2A)=2sinA*cosA==>(1+sin2A-cos2A)/(1+sin2A+cos2A)=
tan(4分之派+a)=(tan(pi/4)+tana)/(1-tan(pi/4)tana)=(1+tana)/(1-tana)=1/2tana=-1/3(sin2a-cos2a)/(1+cos2a)
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证:∵(sin2a-cos2a)^2=sin²2a+cos²2a-2sin2acos2a而sin²2a+cos²2a=1,2sin2acos2a=sin4a,∴
因为sinA+cosA=1/2(sinA+cosA)^2=1/4sin^2A+cos^2A+2sinA*cosA=1/4因此,有1+2sinA*cosA=1/41+sin2A=1/4sin2A=-3/
tana=[sin2a]/[cos2a+1]=A/(B+1)tan(a+π/4)=[tana+tan(π/4)]/[1-tanatan(π/4)]=[1+tana]/[1-tanA]=(A+B+1)/
分子=sin²a+cos²a+2siacosa-(cos²a-sin²a)=(sina+cosa)²-(cosa+sina)(cosa-sina)=(
sin2a=cos2a*2=2cos2acos2a=-sin2a*2=-22sin2a
(sin2a-cos2a)^2=sin2a^2+cos2a^2-2*sin2a*cos2a=1-2*sin2a*cos2a=1-sin4a因为sin2a^2+cos2a^2=12*sin2a*cos2
sin2a=2sina*cosacos2a=cos²a-sin²a=2cos²a-1=1-2sin²a所以(1+cos2a)/sin2a-sin2a/(1-co
因为cos2a=B,即cos^2a-sin^2a=B(^2表示平方)所以1-2sin^2a=B化简得:sina=(|2-2B)/2(|表示根号)(1)由sin2a=A得sina=A/2cosa(2)(