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