已知tana=1,sin(2a β)=3sinβ,则tan(a β)=
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sin(2A+B)=sin(2A)cosB+cos(2A)sinB,sin(2A)=2tanA/(1+tanA^2)=1,cos(2A)=(1-tanA^2)/(1+tanA^2)=0,则由sin(2
tana=sina/cosa=1/2,所以cosa=2sina,所以根据sin^2a+cos^2a=1,求出sin,cos
2.2,画个三角就出来了
先不管-1.前面两项除以cos²+sin²,分子分母约掉cos²,代入tan=3.我算得-0.1
sin^2a+sina+2=2sinacosa+sina+2sina/cosa=1/2sin^2a+cos^a=1解出sina和cosa后代入前面的式子就行了.
tana/(tana-1)=-1,tana=1/2sin²a+sinacosa+2=(sin²a+sinacosa)/1+2=(sin²a+sinacosa)/(sin^
tan(A-B)=(tanA-tanB)/(1+tanA*tanB)tan(A-B)/tanA+sin²C/sin²A=1左右移项得1-[(tanA-tanB)/(1+tanA*t
由sin(a+b)=sinacosb+cosasinbsin(a+b)=sinacosb+cosasinb可知sinacosb=[sin(a+b)+sin(a-b)]/2=5/24cosasinb=[
tana=根号2;sin²a=4/5,cos²a=1/52sin²a-sinacosa+cos²a-1=sin²a-sinacosa=sin²
1sin(a+b)=2/3sina*cosb+cosa*sinb=2/3-----------(1)sin(a-b)=1/5sina*cosb-cosa*sinb=1/5-----------(2)联
分析法倒推tanr=-tan(a-r-a)=[tana-tan(a-r)]/[1+tana*tan(a-r)]tana*tanr=[tan^2a-tana*tan(a-r)]/[1+tana*tan(
3*sin(a+b-a)=sin(a+b+a);3*sin(a+b)cos(a)-3*cos(a+b)sin(a)=sin(a+b)cos(a)+cos(a+b)sin(a);2*sin(a+b)co
tana=1a=kπ+π/43sinB=sin(2a+B)=sin(2kπ+π/2+B)=cosB3sinB=cosBsin^2B+cos^2B=1cos^2B=9/10sin^2B=1/10sinB
别做了没用、以后生活也用不着
sin平方2a+sin2acos2a-cos2a=1sin平方2a-1+cos2a(2sinacosa-1)=0-(cos2a)^2+cos2a(2sinacosa-1)=0cos2a(2sinaco
(1+tana)/(1-tana)=3+2√2化简求得tana=(根号2)/2所以cosa=2/根号6sina=(根号2)/(根号6)cos(π-a)方+sin(π+a)cos(π-a)+sin(a-
∵sin²(2a)+sin(2a)cosa-cos(2a)=1∴sin²(2a)+sin(2a)cosa-cos(2a)-1=0==>sin²(2a)+sin(2a)co
sin(2π+a)cos(-π+a)/cos(-a)tana=sin(a)cos(π-a)/cos(a)(sina/cosa)=-sinacosa/sina=-cosa=-1/4
sin^2a*tana+cos^2a*1/tana+2sina*cosa=(1-cos^2a)*tana+(1-sin^2a)*1/tana+2sina*cosa=tana-sina*cosa+1/t
sin^2a+sinacosa/sin^2+1?是sin^2a+sinacosa/sin^2a+1吧?sin^2a+sinacosa/sin^2a+1=sin^2a+sinacosa/(2sin^2a