Maths: Make dn Subject in Equation

[questions_uk]

How can dn be made the subject in the following equation?


It = Io sin2 x 4 x theta x sin2 (pi x dn x p / lambda)

Thanks

 

[HallsofIvy]

$I_t= I_0 sin(2) 4\theta sin^2(\pi d_n p/\lambda)$?
I am assuming that the first sin2 is sin(2) rather than $sin^2$ because sin (and so $sin^2$ is a <b>function</b> not a number and does not make sense if not applied to some number.

You start just the way I'm sure you have learned before: since the $d_n$ you want to solve for is inside the $sin^2$ you first divide both sides by every thing multiplying that: [itex]\frac{I_t}{4 sin(2)\theta} = sin^2(\pi dn p/\lambda)<br /> <br /> Now get rid of the square by doing the opposite of that: take the square root of each side to get $\sqrt{\frac{I_t}{4 sin(2)\theta}}= sin(\pi dn p/\lamba)$<br /> <br /> Now get rid of the sin by doing <b>its</b> opposite: arcsine or $sin^{-1}$ (which is NOT sin to the negative one power!):<br /> $sin^{-1}(\sqrt{\frac{I_t}{4 sin(2)\theta}})= \pi dn p/\lambda$<br /> <br /> Finally, multiply both sides by $\lambda$ and divide both sides by $\pi p$.[/itex]

 

[questions_uk]

$I_t= I_0 sin(2) 4\theta sin^2(\pi d_n p/\lambda)$?
I am assuming that the first sin2 is sin(2) rather than $sin^2$ because sin (and so $sin^2$ is a <b>function</b> not a number and does not make sense if not applied to some number.

You start just the way I'm sure you have learned before: since the $d_n$ you want to solve for is inside the $sin^2$ you first divide both sides by every thing multiplying that: [itex]\frac{I_t}{4 sin(2)\theta} = sin^2(\pi dn p/\lambda)<br /> <br /> Now get rid of the square by doing the opposite of that: take the square root of each side to get $\sqrt{\frac{I_t}{4 sin(2)\theta}}= sin(\pi dn p/\lamba)$<br /> <br /> Now get rid of the sin by doing <b>its</b> opposite: arcsine or $sin^{-1}$ (which is NOT sin to the negative one power!):<br /> $sin^{-1}(\sqrt{\frac{I_t}{4 sin(2)\theta}})= \pi dn p/\lambda$<br /> <br /> Finally, multiply both sides by $\lambda$ and divide both sides by $\pi p$.[/itex]

$<br /> Thanks for your reply.<br /> <br /> Both sins are sin ^ 2. Would this make a difference?$

 

[CRGreathouse]

Thanks for your reply.

Both sins are sin ^ 2. Would this make a difference?

In that case, we don't understand the problem. The following string of symbols:
$$I_t= I_0 sin^2()\cdot 4\theta sin^2(\pi d_n p/\lambda)$$
does not form a well-defined equation because there's nothing in the (). What did you actually mean? Perhaps
$$I_t= I_0 sin^2\left(4\theta sin^2(\pi d_n p/\lambda)\right)$$?

 

[questions_uk]

In that case, we don't understand the problem. The following string of symbols:
$$I_t= I_0 sin^2()\cdot 4\theta sin^2(\pi d_n p/\lambda)$$
does not form a well-defined equation because there's nothing in the (). What did you actually mean? Perhaps
$$I_t= I_0 sin^2\left(4\theta sin^2(\pi d_n p/\lambda)\right)$$?

thanks for the reply. it is (4 x theta) sin ^ 2

 

[HallsofIvy]

Thanks for your reply.

Both sins are sin ^ 2. Would this make a difference?

You mean it is really It= Iosin^2(4 theta) sin^2(pi dn p/lambda)? Even easier. First divide both sides by Iosin^2(4 theta) to get
It/(Io sin^2(4 theta))= sin^2(pi dn p/lambda)

Take square roots:
sqrt(It/(Io sin^2(4 theta)))= sin(pi dn p/lambda)
Take the inverse sine of each side
arcsin(sqrt(It/(Io sin^2(4 theta)))= pi dn p/lambda
and finally, divide both sides by pi d/lambda

lambda arcsin(sqrt(It/Io sin^2(4 theta)))/(pi p)= dn.

 

[questions_uk]

Hi. Thanks for the previous input. how would theta be made the subject? any input is appreciated.