amaths三角函數

tanθ = 2ab/a^2+b^2 or tanθ = 2ab/(a^2-b^2)

2009-03-16 03:51:52 補充：
find the general solution of the equation
cos x - √2 cos 2x + cos 3x =0
cosx+cos3x-√2 cos 2x=0
cos(2x-x)+cos(2x+x)-√2 cos 2x=0
2cos2xcosx-√2 cos 2x=0
√2 cos2x(√2 cos x-1)=0
so cos2x=0 or cosx=√2 /2
2x=2n π or x=2n π- π/4
x=nπ or x=nπ- π/8

2.If tanθ = 2ab/(a^2-b^2) and 0<θ<90 ,then cosθ = ?
tanθ = 2ab/(a^2-b^2) and 0<θ<90
(a^2-b^2)^2+(2ab)^2
=a^4-2a^2b^2+b^4+4a^2b^2
=a^4+2a^2b^2+b^4
=(a^2+b^2)^2
so cosθ=|2ab|/(a^2+b^2)
we do"nt know ab is positive or negative

2.If tanθ = 2ab/(a^2+b^2) and 0<θ<90 ,then cosθ = ?
tanθ = 2ab/(a^2+b^2) and 0<θ<90
(a^2+b^2)^2+(2ab)^2
=a^4+2a^2b^2+b^4+4a^2b^2
=a^4+6a^2b^2+b^4
so cosθ=|2ab|/(a^4+6a^2b^2 +b^4)^(1/2)

2009-03-16 18:58:40 補充：
2cos2xcosx-√2 cos 2x=0
不明這步
√2 cos2x(√2 cos x-1)=0

2cos2xcosx-√2 cos 2x=0
√2 (√2 cos2xcosx-cos2x)=0 提出√2
√2cos2x (√2cosx-1)=0　　提出cos2x

2009-03-17 09:54:46 補充：
√2*√2=2