化简:1)1+sin2θ-cos2θ/1+sin2θ+cos2θ

1个回答

  • 1)1+sin2θ-cos2θ/1+sin2θ+cos2θ

    =(1-cos2θ)+sin2θ/(1+cos2θ)+sin2θ

    =2(sinθ)^2+2sinθcosθ/2(cosθ)^2+2sinθcosθ

    =sinθ/cosθ

    =tanθ

    2)sin(a+π/4)sin(π/4-a)=

    cos(π/2-a-π/4)sin(π/4-a)=

    cos(π/4-a)sin(π/4-a)=

    sin(π/2-2a)/2=

    cos2a/2

    3)sin50(1+√3tan10)

    =(2sin40sin50)/cos10

    =[cos(50-40)-cos(50+40)]/cos10

    =cos10/cos10

    =1

    1)sinX+cosX=1/5 ,sinX-1/5=-cosX ,sin^2X-(2/5)sinX+1/25=cos^2X=1-sin^2X

    2sin^2X-(2/5)sinX-24/25=0,(2sinX+6/5)(sinX-4/5) ,sinX=4/5,(sinX=-3/5舍去)

    cosX=-3/5 ,tanX=-4/3

    2)sin7°=sin(15°-8°)=sin15°cos8°-cos15°sin8°

    cos7°=cos(15°-8)=cos8°cos15°+sin8°sin15°

    所以,问题就简单了,把上面写的那个带进去之后

    得到sin15°cos8°/cos8°cos15°=tan15°

    3)cosA=cos(A-π/6+π/6)=cos(A-π/6)cosπ/6-sin(A-π/6)sinπ/6

    0<A<π/2

    -π/6<A-π/6<π/3

    cos(A-π/6)>0

    sin(A-π/6)=1/3

    cos(A-π/6)=(2√2)/3

    cosA=(2√2)/3*√3/2-(1/3)*(1/2)

    =√6/3-(1/6)

    =[(2√6)-1]/6

    1)原式=(1/2sinacosa+cos^2 a)/(sin^2 a+cos^2 a) 上下同时除以cos^2 a得 (1/2tana+1)/(tan^2 a+1)=3/4

    2)【解】两边平方

    (3sinA+4cosB)^2=36

    得9sin^2A +16cos^2B +24sinAcosB=36 ①

    (4sinB+3cosA)^2=1

    得16sin^2B +9cos^2A +24sinBcosA=1 ②

    ①+ ②

    得:(9sin^2A +9cos^2A) +(16cos^2B+ 16sin^2B) +24sinAcosB+24sinBcosA=37

    即 9+16+24sin(A+B)=37

    所以sin(A+B)=1/2,

    所以A+B=5π/6 或者π/6

    若A+B=π/6,则cosA>√3/2

    3cosA>3√3/2>1 ,则4sinB+3cosA>1 这是不可能的

    所以A+B=5π/6

    因为A+B+C=180

    所以 C=π/6

    3)cosa=4/5,cos(a+b)=cosacosb-sinasinb=4/5cosb-3/5sinb=-4/5(1),又0