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tiny-tim
#2
Dec26-09, 04:57 PM
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Hi ╔(σ_σ)╝!

(have a theta: θ and a phi: φ )
Quote Quote by ╔(σ_σ)╝ View Post
I found that
P dot Q = cos([tex]\phi[/tex] - [tex]\theta[/tex])


But the answer at the back of my book says
cos([tex]\phi[/tex])cos( [tex]\theta[/tex]) - sin([tex]\phi[/tex])sin( [tex]\theta[/tex])
I guess this is the same P dot Q, so my answer is the same as the book.
If that's meant to be cos(θ - φ), then the book is wrong.
3) So |P - Q|/2 = 1-cos[tex]\varphi[/tex]

Right ?

But then my book says the answer is |sin (0.5([tex]\phi[/tex] - [tex]\theta[/tex]))|
i] |A| is the square-root of A.A

ii] 1 + cos2θ = 2sin2θ