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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
The equation for making theta the subject from the expression It = Io sin²(4θ) sin²(πdn p / λ) is derived through a series of algebraic manipulations. First, isolate sin²(4θ) by dividing both sides by Io sin²(πdn p / λ), resulting in sin²(4θ) = It / (Io sin²(πdn p / λ)). Next, take the square root to obtain sin(4θ) = √(It / (Io sin²(πdn p / λ))). Finally, apply the arcsin function to isolate 4θ, leading to 4θ = arcsin(√(It / (Io sin²(πdn p / λ))) and divide by 4 to solve for θ.
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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 sin^2(4\theta) is a constanat so just divide both sides by Io sin^2(\pi dn p/\lambda to get
sin^2(4\theta)= \frac{In}{Iosin^2(\pi dn p/\lambda)}
Get rid of the "2" by doing the opposite: square root
sin(4\theta)= \sqrt{\frac{In}{Iosin^2(\pi dn p/\lambda)}
and get rid of the sin by using arcsin:
4\theta= arcsin\left(\sqrt{\frac{In}{Iosin^2(\pi dn p/\lambda}\right)
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!