Need help with alternative solution to physics problem.

Question

I am trying to solve a physics problem (in a different way then the solution in the book because its more straightforward, answer was approximately $56.5$ degrees), and during that process, I got an expression I have to solve. I am pretty sure there isn't an analytic solution so I just want a numerical answer. I went through different websites and none of them can solve it, mainly because the expression is too long.

I want to find the value of $\theta$ that satisfies this. (By the way, the expression in the text equals $0$)

P.S: The $-$ before the fractions is a negative sign, $0\leq\theta\leq \frac{\pi}{2}$ and $v,g > 0$.

$$-\dfrac{2v^2\cos\left({\theta}\right)\sin\left({\theta}\right)\operatorname{arsinh}\left(\frac{\sec^2\left({\theta}\right)\sin\left(2{\theta}\right)}{2}\right)}{g}\\ +\dfrac{v^2\cos^2\left({\theta}\right)\left(\sec^2\left({\theta}\right)\tan\left({\theta}\right)\sin\left(2{\theta}\right)+\sec^2\left({\theta}\right)\cos\left(2{\theta}\right)\right)}{g\sqrt{\frac{\sec^4\left({\theta}\right)\sin^2\left(2{\theta}\right)}{4}+1}}\\ -\dfrac{v^2\sin^2\left(2{\theta}\right)}{2^\frac{3}{2}g\cos^2\left({\theta}\right)\sqrt{\cos\left(2{\theta}\right)+1}}\\ +\dfrac{v^2\sin\left({\theta}\right)\sqrt{\cos\left(2{\theta}\right)+1}\sin\left(2{\theta}\right)}{\sqrt{2}\,g\cos^3\left({\theta}\right)}\\ +\dfrac{v^2\cos\left(2{\theta}\right)\sqrt{\cos\left(2{\theta}\right)+1}}{\sqrt{2}\,g\cos^2\left({\theta}\right)}$$

Answer 1

I am going to tackle this by breaking the expression down into parts and simplifying the parts one by one before putting it back together.

Section 1: $\frac{v^2\sin(2\theta)\sin\theta\sqrt{\cos(2\theta)+1}}{g\cos^3(\theta)\sqrt{2}}+\frac{v^2\cos(2\theta)\sqrt{\cos(2\theta)+1}}{g\cos^2(\theta)\sqrt{2}}$

Now, multiplying the second fraction by $\cos\theta$ and adding the numerators, we get:

$\frac{(v^2\sqrt{\cos(2\theta)+1})(\sin2\theta\sin\theta+\cos2\theta\cos\theta)}{g\cos^3(\theta)\sqrt{2}}$

Looking at $\sin2\theta\sin\theta+\cos2\theta\cos\theta$, we can expand $\sin2\theta$ and $\cos2\theta$ before simplifying it.

Doing so, we get $\cos\theta$. Plugging it back into our simplified Section $1$ expression and simplifying further, we get:

$\frac{v^2\sqrt{\cos(2\theta)+1}}{g\cos^2(\theta)\sqrt{2}}$

Now, looking at $\sqrt{\cos2\theta + 1}$, using the $\cos2\theta$ identity for $\cos\theta$, we get: $\cos\theta\sqrt{2}$.

Plugging it into the expression we got above, we get:

$\frac{v^2}{g\cos\theta}$

Therefore, our final expression for Section 1 is $\fbox{$\frac{v^2}{g\cos\theta}$}$.

Section 2: $\frac{v^2\cos^2(\theta)(\sec^2(\theta)\tan\theta\sin2\theta + \sec^2(\theta)\cos2\theta)}{g\sqrt{\frac{\sec^4(\theta)\sin^2(2\theta)}{4}+1}}$

Looking at $\sec^4(\theta)\sin^2(2\theta)$, we know that $\sec\theta = \frac{1}{\cos\theta}$ and that $\sin2\theta = 2\sin\theta\cos\theta$.

Using these facts, $\sec^4(\theta)\sin^2(2\theta) = 4\tan^2(\theta)$

Plugging it back into the denominator of Section 2's expression, we get:

$\frac{v^2\cos^2(\theta)(\sec^2(\theta)\tan\theta\sin2\theta + \sec^2(\theta)\cos2\theta)}{g\sqrt{\frac{4\tan^2(\theta)}{4}+1}}$

Looking at the denominator of the above expression, this can be simplified into $g\sec\theta$.

Plugging this back into the above expression and rearranging, we get:

$\frac{v^2\cos^3(\theta)(\sec^2(\theta)\tan\theta\sin2\theta + \sec^2(\theta)\cos2\theta)}{g}$

Looking at $\sec^2(\theta)\tan\theta\sin2\theta + \sec^2(\theta)\cos2\theta$, we can use the fact that $\sec^2(\theta)=\frac{1}{\cos^2(\theta)}$ and utilise the double angle formulae for $\sin2\theta$ and $\cos2\theta$.

Doing so, we get $\sec^2(\theta)$.

Plugging it back into the Section 2 expression above and simplifying, we get:

$\frac{v^2\cos(\theta)}{g}$.

Therefore, our final expression for Section 2 is $\fbox{$\frac{v^2\cos(\theta)}{g}$}$.

Looking at both expressions from Sections 1 and 2, we can add them together.

Doing so, we get:

$\frac{v^2}{g}(\cos\theta + \sec\theta)$

Since $\cos\theta + \sec\theta$ cannot be simplified further, $\fbox{$\frac{v^2}{g}(\cos\theta + \sec\theta)$}$ is our final combined expression from Sections 1 and 2.

Section 3: $-\frac{v^2\sin^2(2\theta)}{2^{\frac{3}{2}}g\cos^2(\theta)\sqrt{\cos2\theta+1}}$

Remembering that $\sqrt{\cos2\theta+1} = \cos\theta\sqrt{2}$, the expression for section 3 can be reduced to the following:

$-\frac{v^2\sin^2(2\theta)}{4g\cos^3(\theta)}$

Remembering that $\sin2\theta=2\sin\theta\cos\theta$, we can simplify the above expression even further to

$\fbox{$-\frac{v^2\tan\theta\sin\theta}{g}$}$

Putting this into our combined Sections 1 and 2 expression, we get:

$\frac{v^2}{g}(\cos\theta + \sec\theta - \tan\theta\sin\theta)$

Section 4: $-\frac{2v^2\sin\theta\cos\theta\operatorname{arsinh}\left(\frac{\sec^2\left({\theta}\right)\sin\left(2{\theta}\right)}{2}\right)}{g}$

Remembering that $2\sin\theta\cos\theta = \sin2\theta$ and that $\sec^2(\theta)=\frac{1}{\cos^2(\theta)}$, we can simplify the Section 4 expression into the following:

$\fbox{$-\frac{v^2\sin2\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right)}{g}$}$

Now, combining our Sections 1, 2 and 3 expression and our simplified Section 4 expression, we get:

$\frac{v^2}{g}(\cos\theta+\sec\theta-\tan\theta\sin\theta-\sin2\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right))$

Let's call the above the final expression (basically the original expression but simplified).

Final Expression:$\frac{v^2}{g}(\cos\theta+\sec\theta-\tan\theta\sin\theta-\sin2\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right))$

Looking at $\cos\theta+\sec\theta-\tan\theta\sin\theta-\sin2\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right)$, we should expand everything before re-examining the expression.

Doing so gives us: $\cos\theta+\sec\theta-\sin^2(\theta)\sec\theta-2\sin\theta\cos\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right)$

Looking at $\sec\theta -\sin^2(\theta)\sec\theta$, we can factor out $\sec\theta$ and simplify. Doing so gives us $\cos\theta$.

Plugging this back into our expression gives us:

$2\cos\theta-2\sin\theta\cos\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right)$

Taking out $2\cos\theta$, we get $2\cos\theta(1-\sin\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right))$

Plugging this back into our final expression, we get:

$\frac{v^2}{g}(2\cos\theta)(1-\sin\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right))$

Therefore, our final form for the final expression is $\fbox{$\frac{2v^2}{g}\cos\theta(1-\sin\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right))$}$

Numerical Answer to Problem

Now that we have turned the original expression into the final expression, our last step is to equate the final expression to $0$ and see which $\theta$ satisfies the final expression (remembering that $0 \leq \theta \leq \frac{\pi}{2}$ and that $v,g \geq 0$).

Doing so, we get:

$\frac{2v^2}{g}\cos\theta(1-\sin\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right))=0$

Multiplying both sides by $\frac{g}{2v^2}$ before manipulating the expression on the LHS gives us:

$1=\sin\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right))$

Now the best way to solve this equation is to look at the graph of $\sin\theta\operatorname{arsinh}\left(\tan\left({\theta}\right)\right))$, which is pictured below:

If we then drew the line $y=1$ across this graph, we would get the following:

Now, since $0 \leq \theta \leq \frac{\pi}{2}$, we will only be looking at the positive solution.

The positive solution is $\theta \approx 0.985515$ radians (to $6$ d.p.)

Multiplying this by $\frac{180}{\pi}$, the answer to this problem (in degrees) is approximately $\fbox{$56.46^{\circ}$}$ (to $2$ d.p.)

Note 1: For anybody wondering how did I get certain results in this solution, I had pen and paper by my side and worked it out first before showing whatever I deemed was neccessary.

Answer 2

This is an extended comment to help verify @Yajat's answer.

---

We can write the expression as

$$\frac{v^2}{g}\left( \begin{align} -2\cos t\sin t \operatorname{arsinh}p +\frac{r}{\sqrt{p^2+1}} -\frac{\sin^22t}{2\sqrt{2}\,q\cos^2t} +\frac{q\sin t\sin2t}{\sqrt{2}\,\cos^3t} +\frac{q\cos2t}{\sqrt{2}\,\cos^2t} \end{align}\right) \tag 1$$

where

$$\begin{align} p &:= \frac12\sec^2t\sin2t=\frac12\cdot\frac{1}{\cos^2t}\cdot2\sin t\cos t = \frac{\sin t}{\cos t} =\tan t \tag2\\[8pt] \sqrt{p^2+1}&=|\sec t| \tag3\\[8pt] q &:= \sqrt{\cos2t+1} =\sqrt{2}\,|\cos t| \tag4\\[8pt] r &:= \cos^2t\left(\sec^2t\tan t\sin2t+\sec^2t\cos2t\right) = \tan t\sin2t+\cos2t \\[4pt] &= \frac{\sin t}{\cos t}\cdot 2\sin t \cos t + 1 - 2\sin^2 t = 1 \tag5 \end{align}$$

Noting that we can drop the absolute values on $\cos t$ and $\sec t$ for $0\leq t\leq \pi/2$, we substitute the simplifications into the last four terms of $(1)$ to get $$\begin{align} &\frac{1}{\sec t} -\frac{\sin^22t}{2\sqrt{2}\cdot \sqrt{2}\cos t\cdot\cos^2t} +\frac{\sqrt{2}\cos t\cdot\sin t\sin2t}{\sqrt{2}\,\cos^3t} +\frac{\sqrt{2}\cos t\cdot\cos2t}{\sqrt{2}\,\cos^2t} \tag6\\[8pt] =\quad&\cos t -\frac{4\sin^2t\cos^2t}{4\cos^3t} +\frac{\sin t\cdot 2\sin t\cos t}{\cos^2t} +\frac{1-2\sin^2t}{\cos t} \tag7\\[8pt] =\quad&\frac1{\cos t}\left( \cos^2 t -\sin^2t +2\sin^2 t +1-2\sin^2t \right) \tag8\\[8pt] =\quad&2\cos t \tag9 \end{align}$$

Thus, $(1)$ becomes $$\frac{2v^2}{g}\cos t\left(\,1-\sin t\operatorname{arsinh}(\tan t)\,\right) \tag{$\star$}$$