- #1
SUDOnym
- 90
- 1
I always get muddled when I'm dealing with chain rule of any degree of complexity and also when dealing with powers of trig. functions - this problem contains both:
find [tex]\frac{\partial n}{\partial A}[/tex] and [tex]\frac{\partial n}{\partial D}[/tex] of the following function:
[tex]n=\frac{\sin(\frac{A+D}{2})}{\sin(\frac{A}{2})}[/tex]
also find [tex]|\frac{\partial n}{\partial D}|^{2}[/tex] [tex]|\frac{\partial n}{\partial A}|^{2}[/tex]
My Attempt:
[tex]n=\frac{\sin(\frac{A+D}{2})}{\sin(\frac{A}{2})}[/tex]
[tex]\implies n=\sin(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-1}[/tex]
[tex]\implies\frac{\partial n}{\partial A}=\frac{1}{2}\cos(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-1}+\sin(\frac{A+D}{2})(-1)[\sin(\frac{A}{2})]^{-2}(\frac{1}{2})\cos(\frac{A}{2})[/tex]
[tex]=\frac{1}{2}\frac{\cos(\frac{A+D}{2})}{\sin(\frac{A}{2})}-\frac{1}{2}\frac{\sin(\frac{A+D}{2})\cos(\frac{A}{2})}{[\sin(\frac{A}{2})]^{2}}[/tex]
[tex]=\frac{1}{2\sin(\frac{A}{2})}[\cos(\frac{A+D}{2})-\sin(\frac{A+D}{2})\arctan(\frac{A}{2})][/tex]
[tex]\implies|\frac{\partial n}{\partial A}|^{2}=\frac{1}{4\sin^{2}(\frac{A}{2})}[\cos^{2}(\frac{A+D}{2})+\sin^{2}(\frac{A+D}{2})\arctan^{2}(\frac{A}{2})-2\cos(\frac{A+D}{2})\sin(\frac{A+D}{2})\arctan(\frac{A}{2})][/tex]
*Correction that last arctan term should read [tex]\arctan(\frac{A}{2})[/tex]
At this point I tried to simplify the trig. a bit... (but a don't think I did a great job!):
[tex]=\frac{1}{4\sin^{2}(\frac{A}{2})}[\frac{1}{2}\cos(A+D)+1]+[\frac{1}{2}(-\cos(A+D)+1)]\arctan^{2}(\frac{A}{2})-\sin(A+D)\arctan(\frac{A}{2})[/tex]
Main issues I have with the above are: I am never any good with chain rule and also re. the trig. I keep getting confused as to whether for example the function:
[tex]\sin^{x}(A)[/tex]
is in all cases completely equivalent to:
[tex][\sin(A)]^{x}[/tex]
ie. are there some cases where they are not equivalent? - this confusion may or may not have caused errors in my attempt above...
OK, now [tex]\frac{\partial n}{\partial D}[/tex] is a little easier:
[tex]\frac{\partial n}{\partial D}=\frac{1}{2}\cos(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-1}[/tex]
[tex]\implies|\frac{\partial n}{\partial D}|^{2}=\frac{1}{4}\cos^{2}(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-2}[/tex]All advice and corrections are greatly appreciated.
find [tex]\frac{\partial n}{\partial A}[/tex] and [tex]\frac{\partial n}{\partial D}[/tex] of the following function:
[tex]n=\frac{\sin(\frac{A+D}{2})}{\sin(\frac{A}{2})}[/tex]
also find [tex]|\frac{\partial n}{\partial D}|^{2}[/tex] [tex]|\frac{\partial n}{\partial A}|^{2}[/tex]
My Attempt:
[tex]n=\frac{\sin(\frac{A+D}{2})}{\sin(\frac{A}{2})}[/tex]
[tex]\implies n=\sin(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-1}[/tex]
[tex]\implies\frac{\partial n}{\partial A}=\frac{1}{2}\cos(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-1}+\sin(\frac{A+D}{2})(-1)[\sin(\frac{A}{2})]^{-2}(\frac{1}{2})\cos(\frac{A}{2})[/tex]
[tex]=\frac{1}{2}\frac{\cos(\frac{A+D}{2})}{\sin(\frac{A}{2})}-\frac{1}{2}\frac{\sin(\frac{A+D}{2})\cos(\frac{A}{2})}{[\sin(\frac{A}{2})]^{2}}[/tex]
[tex]=\frac{1}{2\sin(\frac{A}{2})}[\cos(\frac{A+D}{2})-\sin(\frac{A+D}{2})\arctan(\frac{A}{2})][/tex]
[tex]\implies|\frac{\partial n}{\partial A}|^{2}=\frac{1}{4\sin^{2}(\frac{A}{2})}[\cos^{2}(\frac{A+D}{2})+\sin^{2}(\frac{A+D}{2})\arctan^{2}(\frac{A}{2})-2\cos(\frac{A+D}{2})\sin(\frac{A+D}{2})\arctan(\frac{A}{2})][/tex]
*Correction that last arctan term should read [tex]\arctan(\frac{A}{2})[/tex]
At this point I tried to simplify the trig. a bit... (but a don't think I did a great job!):
[tex]=\frac{1}{4\sin^{2}(\frac{A}{2})}[\frac{1}{2}\cos(A+D)+1]+[\frac{1}{2}(-\cos(A+D)+1)]\arctan^{2}(\frac{A}{2})-\sin(A+D)\arctan(\frac{A}{2})[/tex]
Main issues I have with the above are: I am never any good with chain rule and also re. the trig. I keep getting confused as to whether for example the function:
[tex]\sin^{x}(A)[/tex]
is in all cases completely equivalent to:
[tex][\sin(A)]^{x}[/tex]
ie. are there some cases where they are not equivalent? - this confusion may or may not have caused errors in my attempt above...
OK, now [tex]\frac{\partial n}{\partial D}[/tex] is a little easier:
[tex]\frac{\partial n}{\partial D}=\frac{1}{2}\cos(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-1}[/tex]
[tex]\implies|\frac{\partial n}{\partial D}|^{2}=\frac{1}{4}\cos^{2}(\frac{A+D}{2})[\sin(\frac{A}{2})]^{-2}[/tex]All advice and corrections are greatly appreciated.
Last edited: