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francisg3
Jan25-11, 09:40 PM
This is part of a bigger problem, I need to find an expression in terms of θ for the expression below.


1. x=(2r sin⁡θ)/3θ where x=r/2
2. r/2=(2r sin⁡θ)/3θ
3. r=(4r sin⁡θ)/3θ
4. 1=(4 sin⁡θ)/3θ
5. 1/4= sin⁡θ/3θ
6. 3θ/4= sin^-1(θ)

I don't know where to go from there. All I need is an expression in the form of θ= ...


Thanks!

tiny-tim
Jan26-11, 09:41 AM
hi francisg3! :smile:

(try using the X2 icon just above the Reply box :wink:)
1. x=(2r sin⁡θ)/3θ where x=r/2
2. r/2=(2r sin⁡θ)/3θ
3. r=(4r sin⁡θ)/3θ
4. 1=(4 sin⁡θ)/3θ
5. 1/4= sin⁡θ/3θ
6. 3θ/4= sin^-1(θ)

I don't know where to go from there. All I need is an expression in the form of θ= ...


Thanks!

i'm confused :redface: …

6. has sin-1θ …

were they all supposed to have sin-1θ ? :confused:

anyway, when you have A = sin-1B,

you can take sin of both sides, giving you sinA = B :wink:

HallsofIvy
Jan26-11, 10:25 AM
However, in general, if you have the unknown both inside a transcendental function, such as sine) and outside it, there is no simple "algebraic" way to solve for it.