questions_uk
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hi. with It = Io sin^2 x (4 x theta) x sin^2 (pi x dn x p / lambda) what would the equation be for making theta the subject? thanks for your input
questions_uk said:hi. with It = Io sin^2 x (4 x theta) x sin^2 (pi x dn x p /lambda) what would the equation be for making theta the subject? thanks for your input
HallsofIvy said:questions_uk, start by "reversing" what is there, just like solving any equation. Everything except [itex]sin^2(4\theta)[/itex] is a constanat so just divide both sides by [itex]Io sin^2(\pi dn p/\lambda[/itex] to get
[tex]sin^2(4\theta)= \frac{In}{Iosin^2(\pi dn p/\lambda)}[/tex]
Get rid of the "2" by doing the opposite: square root
[tex]sin(4\theta)= \sqrt{\frac{In}{Iosin^2(\pi dn p/\lambda)}[/tex]
and get rid of the sin by using arcsin:
[tex]4\theta= arcsin\left(\sqrt{\frac{In}{Iosin^2(\pi dn p/\lambda}\right)[/tex]
Finally, of course, divide both sides by 4.
Since squaring and sine are not "one-to-one" functions, you might need to think about other possible values.
Symbolipoint, use "\", not "/" inside LaTex. And I recommend that you put entire equations in LaTex, not just individual symbols. In the second one here, you had a "tex", "/tex" pair inside another!