Transposition of spherical Trig formula

[Steved154]

Good day all,

I'm a teacher (NOT a math teacher) and am having a mind blank here and need help before I loose what little hair I have remaining trying to do this with a class next week.

Cos AB = (Cos P x Sin PA x Sin PB) + (Cos PA x Cos PB)

If I am given the value of AB, P and PB, how can I transpose that formula to find PA? I end up at a point with Cos PA and Sin PA on oneside of the equation.

Appreciate any help.

Regards
Stephen

 

[Dr.Peterson]

Good day all,

I'm a teacher (NOT a math teacher) and am having a mind blank here and need help before I loose what little hair I have remaining trying to do this with a class next week.

Cos AB = (Cos P x Sin PA x Sin PB) + (Cos PA x Cos PB)

If I am given the value of AB, P and PB, how can I transpose that formula to find PA? I end up at a point with Cos PA and Sin PA on one side of the equation.

Appreciate any help.

Regards
Stephen

So, in effect, you want to solve [imath]a\sin(x)+b\cos(x)=c[/imath] for x, right?

To solve this, you can rewrite the combination of sine and cosine as a single (shifted) sine or cosine, using the formula shown here, and explained here.

 

[The Highlander]

Good day all,

I'm a teacher (NOT a math teacher) and am having a mind blank here and need help before I loose what little hair I have remaining trying to do this with a class next week.

Hi Stephen,

If not Maths, I do hope you're not an English teacher; there's only one "o" in lose! lol.
(Sorry, that misspelling is one of my pet hates, )