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this is a problem that im not 100% can be done as i derived the equation myself but please help if you can and if it cant be done let me know:

i need the end product to be $\tan(\theta) = \ldots$ and for there to be no $\theta$ on the RHS. equation is: $$\frac{96.04 \cdot D^2}{\left(U \cdot \cos(\theta)\right)^2} = 2\left(U \cdot \sin(\theta)\right)^2 + (19.6 \cdot W)$$

where D, U and W are constants.

thanks so much in advance!!


i believe i owe you both an apology, it turns out i made a mistake earlier on in my equation so the one i posted was not possible. i only came to this conclusion after subbing in the data values i had into both answers and getting imaginary numbers back. so if i could be a pain, here is the updated (and verified) equation:

\begin{equation*} \frac{D}{U\cos (\theta )}=\frac{U\sin (\theta )}{C}+( \frac{A+U^{2}\sin ^{2}(\theta )}{B}) ^{0.5} \end{equation*}

once again im after $\tan(\theta)=$ etc with no ($\theta$) on the RHS

sorry for the mess around and thanks for helping :)

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Do you mean this? $$\frac{D}{U\cos(\theta)} = \frac{U\sin(\theta)}{C } + (\frac{A + U^2\sin^2(\theta)}{B})^{0.5}$$ – Américo Tavares Feb 11 '14 at 14:59
yes thats the one, sorry i dont know how to use laTex – jordan6194 Feb 12 '14 at 0:43
In the present form this equation is more difficult to solve than before. For some basic information about LaTeX see e.g. here, here, here and here. – Américo Tavares Feb 12 '14 at 1:27
thanks for the links. unfortunately this is the simplest form i can get it to. is it still possible to do or am i wasting my time? – jordan6194 Feb 12 '14 at 4:15
wow yes that got very big very quickly. thankyou for your assistance, thats certainly not something i would like to have arrived at myself. – jordan6194 Feb 12 '14 at 12:07
up vote 0 down vote accepted

UPDATE (answer to the revised question). If we multiply the edited equation by $C\sqrt{B}U\cos \theta \neq 0$, we get the equivalent equation

\begin{equation*} \sqrt{B}CD=\sqrt{B}U^{2}\sin \theta \cos \theta +CU\left( A+U^{2}\sin ^{2}\theta \right) ^{1/2}\cos \theta . \end{equation*}

Using the definition of $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and the fundamental trigonometric identity $\sin ^{2}\theta +\cos ^{2}\theta =1$ , we are able to express $\sin \theta $ and $\cos \theta $, as well as $\sin \theta \cos \theta $, in terms of $t=\tan \theta $. A possible derivation is as follows. From

\begin{equation*} \tan ^{2}\theta =\frac{\sin ^{2}\theta }{\cos ^{2}\theta }=\frac{\sin ^{2}\theta }{1-\sin ^{2}\theta }=\frac{1-\cos ^{2}\theta }{\cos ^{2}\theta }= \frac{1}{\cos ^{2}\theta }-1 \end{equation*}

we obtain successively \begin{eqnarray*} \cos ^{2}\theta &=&\frac{1}{1+\tan ^{2}\theta } \\ &\Rightarrow &\cos \theta =\pm \frac{1}{\sqrt{1+\tan ^{2}\theta }}=\pm \frac{ 1}{\sqrt{1+t^{2}}} \\ && \\ \sin ^{2}\theta &=&\left( 1-\sin ^{2}\theta \right) \tan ^{2}\theta \Leftrightarrow \sin ^{2}\theta =\frac{\tan ^{2}\theta }{1+\tan ^{2}\theta }= \frac{t^{2}}{1+t^{2}} \\ &\Rightarrow &\sin \theta =\pm \frac{\tan \theta }{\sqrt{1+\tan ^{2}\theta }} =\pm \frac{t}{\sqrt{1+t^{2}}} \end{eqnarray*}


\begin{eqnarray*} \sin \theta &=&\cos \theta \tan \theta \\ &\Leftrightarrow &\sin \theta \cos \theta =\cos ^{2}\theta \tan \theta = \frac{\tan \theta }{1+\tan ^{2}\theta }=\frac{t}{1+t^{2}}. \end{eqnarray*}

Substituting these expressions in the first equation yields \begin{eqnarray*} \sqrt{B}CD &=&\frac{\sqrt{B}U^{2}t}{1+t^{2}}+CU\left( A+\frac{U^{2}t^{2}}{ 1+t^{2}}\right) ^{1/2}\left( \pm \frac{1}{\sqrt{1+t^{2}}}\right) \\ \sqrt{B}CD\left( 1+t^{2}\right) -\sqrt{B}U^{2}t &=&\pm CU\left( A+\left( A+U^{2}\right) t^{2}\right) ^{1/2} \end{eqnarray*}

Squaring now both sides and rearranging the terms we obtain the following quartic equation in $t$

\begin{equation*} at^{4}+bt^{3}+ct^{2}+dt+e=0 \end{equation*} whose coefficients are

\begin{eqnarray*} a &=&BC^{2}D^{2} \\ b &=&-2BCDU^{2} \\ c &=&2BC^{2}D^{2}-C^{2}U^{4}+BU^{4}-AC^{2}U^{2} \\ d &=&-2BCDU^{2} \\ e &=&-2BCDU^{2}-AC^{2}U^{2}+BC^{2}D^{2}. \end{eqnarray*}

unfortunately this is the simplest form i can get it to. is it still possible to do or am i wasting my time?

This equation (after depressed) although solvable algebraically by an auxiliary cubic equation has huge solutions (see Wikipedia entry Ferrari's solution).

Multiplying your equation by $U^{2}\cos ^{2}\theta\ne 0 $ we get the equivalent equation

\begin{equation*} 2U^{4}\sin ^{2}\theta \cos ^{2}\theta +19.6WU^{2}\cos ^{2}\theta -96.04D^{2}=0. \end{equation*}

Using the identity $\sin ^{2}\theta =1-\cos ^{2}\theta $ and rearranging the terms we obtain the following quadratic equation in $y=\cos ^{2}\theta $:

\begin{equation*} ay^{2}+by+c=0\Leftrightarrow y=\frac{1}{2a}\left( -b\pm \sqrt{b^{2}-4ac} \right) , \end{equation*}

whose coefficients are

\begin{equation*} a=-2U^{4},\quad b=2U^{4}+19.6WU^{2},\quad c=-96.04D^{2}. \end{equation*}

To get the result we just need to express $\tan \theta $ in terms of $y$

\begin{equation*} \sin ^{2}\theta +\cos ^{2}\theta =1\Leftrightarrow \tan ^{2}\theta +1=\frac{1 }{\cos ^{2}\theta }=\frac{1}{y}\Leftrightarrow \tan \theta =\pm \sqrt{\frac{1}{y}-1}. \end{equation*}

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If $\displaystyle\tan\theta=u,$



or $\displaystyle\sin^2\theta=1-\cos^2\theta=1-\frac1{1+u^2}=\frac{u^2}{1+u^2}$

Rearrange the given relation to form a Quadratic Equation in $u=\tan\theta$

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so its definitely possible to reduce this all down to tan(theta)=... ? because im not having a whole lot of luck at the moment – jordan6194 Feb 9 '14 at 15:08
@jordan6194, in fact as $$\tan^2\theta=$$ – lab bhattacharjee Feb 9 '14 at 15:09
okay thank you ill keep trying, i feel like im making this a lot more complicated than it needs to be ahaha thanks for your help so far – jordan6194 Feb 9 '14 at 15:10
@jordan6194, please simplify and let me know if you are stuck – lab bhattacharjee Feb 9 '14 at 15:13
hoping im on track so far, ive gotten it to here but cant eliminate that last tan in the denominator tan^2(theta).(96.04 . D^2 - (2 . U^4)/(1+tan^2(theta)) = constants. hopefully thats no too confusing to follow – jordan6194 Feb 9 '14 at 15:24

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