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 March 22nd, 2014, 06:11 AM #1 Newbie   Joined: Feb 2011 Posts: 8 Thanks: 0 Help with Cos If $cos h= \frac { A}{ \cos L} - B * \tan L,$ can you help me get the value of L?
March 22nd, 2014, 11:13 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Re: Help with Cos

Hello, moja_a!

This is a truly UGLY problem . . .

Quote:
 $\text{Solve for }x:\;\;\cos H \:=\:\frac { A}{\cos x} \,-\,B\tan x$

$\text{W\!e have: }\:\cos H \:=\:\frac{A}{\cos x}\,-\,B\,\!\frac{\sin x}{\cos x}$

$\text{Multiply by }\cos x:\;\;\cos H\,\!\cos x \;=\;A\,-\,B\,\!\sin x$

$\text{Square both sides: }\;\cos^2H\,\!\cos^2x \;=\;A^2 \,-\,2A\,\!\sin x \,+\,B^2\,\!\sin^2x$

$\;\;\;\cos^2H(1\,-\,\sin^2x) \;=\;A^2\,-\,2AB\,\!\sin x \,+\,B^2\,\!\sin^2x$

$\;\;\;\cos^2H \,-\,\cos^2H\sin^2x \;=\;A^2\,-\,2AB\,\!\sin x \,+\,B^2\,\!\sin^2x$

$\;\;\;(B^2\,+\,\cos^2H)\sin^2x-\,2AB\,\!\sin x \,+\,(A^2\,-\,\cos^2H) \;=\;0$

$\text{Quadratic Formula:}$

$\;\;\;\sin x \;=\;\frac{2AB\,\pm\,\sqrt{4A^2B^2\,-\,4(B^2\,+\,\cos^2H)(A^2\,-\,\cos^2H)}}{2(B^2\,+\,\cos^2H)}$

$\;\;\;\sin x \;=\;\frac{AB\,\pm\,\sqrt{A^2B^2\,-\,(B^2\,+\,\cos^2H)(A^2\,-\,\cos^2H)}}{B^2\,+\,\cos^2H}$

$\;\;\;\sin x \;=\;\frac{AB\,\pm\,\sqrt{\cos^4H\,+\,(B^2\,-\,A^2)\cos^2H}}{B^2\,+\,\cos^2H}$

$\;\;\;\sin x \;=\;\frac{AB\,\pm\,\cos H\sqrt{\cos^2H \,+\,B^2\,-\,A^2}}{B^2\,+\,\cos^2H}$

$\text{Therefore: }\:x \;=\;\arcsin\left(\frac{AB\,\pm\,\cos H\sqrt{\cos^2H \,+\,B^2\,-\,A^2}}{B^2\,+\,\cos^2H}\right)$

 March 22nd, 2014, 11:21 AM #3 Newbie   Joined: Feb 2011 Posts: 8 Thanks: 0 Re: Help with Cos thanks I will try it now

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