What is the equation for making theta the subject in this scenario?

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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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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
 
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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

What are you trying to express? It=Iosin2(4\thetasin2(\pidnp/\lambda)

Maybe some of that needs to be amended as
\frac{\pi dnp}{\lambda}
The typesetting tools available in the Compose for replying are new to me.

I know that did not work, I still have trouble with these typesetting features.
 
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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{It}{Iosin^2(\pi dn p/\lambda)}
Get rid of the "2" by doing the opposite: square root
sin(4\theta)= \sqrt{\frac{It}{Iosin^2(\pi dn p/\lambda)}
and get rid of the sin by using arcsin:
4\theta= arcsin\left(\sqrt{\frac{It}{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!
 
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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!

hi, thanks for the reply. where did the In bit come from?
 
Sorry, that was supposed to be your It.

I have gone back and editted to change In to It.
 
Thank you for your help.
 

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