Trig identity / equation help :) H_c=151.8-sqrt[192584.0-103612.3*...]

[pisoj]

Hello!


I have following equation, from which I would like to obtain X_s variable. I do not care about numerical result of addition / subtraction / division etc, I am only interested in the step by step procedure (if possible), and of course equation that shows X_s variable.


My math skills are a bit (a tad bit more) rusty since I did not do these for a loong time.

Anyway, here is the dude:

[math]\Delta H_c = 151.8 - \sqrt[]{192584.0 - 103612.3 \times \cos(32\degree +\arcsin(\frac{241-X_s} {344})}[/math]




Every help is welcome!



P.S. I am not sure if this is correct forum for this question!

 

[mmm4444bot]

Hello. Your expression contains a mismatched grouping symbol. Is the cosine supposed to be cos(32°) or does the missing close-parenthesis belong after the arcsine?
[imath]\;[/imath]

 

[pisoj]

Hello. Your expression contains a mismatched grouping symbol. Is the cosine supposed to be cos(32°) or does the missing close-parenthesis belong after the arcsine?
[imath]\;[/imath]

Ah, yes my bad. Did not use latex before so I missed it during googling how to write equation

Missing close-parenthesis should go after the arcsine!


[math]ΔH_c=151.8− \sqrt{192584.0−103612.3×cos(32°+arcsin( \frac{241-X_s}{344}))} [/math]

 

[Steven G]

Suppose A = B - sqrt (C+D) and I want to solve for D.

Then sqrt(C+D) = B-A
C+D = (B-A)^2
D = (B-A)^2 -C.

Can you get that far with your problem?

 

[pisoj]

Suppose A = B - sqrt (C+D) and I want to solve for D.

Then sqrt(C+D) = B-A
C+D = (B-A)^2
D = (B-A)^2 -C.

Can you get that far with your problem?

My issue is that I am not sure how to get rid of cos and arcsine :/

 

[mmm4444bot]

My issue is that I am not sure how to get rid of cos and arcsine

Hello. Are you saying that you've already applied Steven's approach and isolated the cosine expression to one side of the equation? (If so, then you ought to have a quadratic expression in H on the other side.) Please show your work or list those steps, so that we may check your progress before proceeding.

Here's a link summarizing the inverse trig functions. Basically, if cos(expression)=output, then arccos(output)=expression. In other words, after you've isolated the cosine expression to one side of the equation, you eliminate the cosine function by applying the arccosine to both sides.

Here's a link to the forum guidelines summary.
[imath]\;[/imath]

 

[Steven G]

Great! Now we know where you are having trouble. Can you please share your work so that we know exactly what help you need?

I will offer this:

A = cos(B + arcsinC), solve for C

Then arccosA = B + arcsinC
arccosA - B = arcsinC
sin(arccosA - B) = C
That is basically what you need to do and clean up my work a bit.