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Please can someone help transpose the following trigonometric equations to make $m_r$ or $\tan\theta_r$ the subject ($m_t$ and $m_r$ are line gradients):

Solve each of the following equations, separately, for $\theta_r$ in terms of $m_t$:

$$m_t=\tan\theta_t=\frac{e-\cos\theta_r}{\sin\theta_r}=e\csc\theta_r-\cot\theta_r\tag{1}$$

$$m_t=\tan\theta_t=\frac{e^2\cos^2\theta_r-\cos^2\theta_r+\sin^2\theta_r}{2\sin\theta_r\cos\theta_r}=e^2\frac{1}{2}\cot\theta_r-\cot2\theta_r \tag{2}$$

These are two independent equations, not a system of equations. For both equations I want to be able to solve:

$$m_r=\tan\theta_r=?$$

I've spent many hours trying but don't have the knowledge :(

Thanks in advance,

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  • $\begingroup$ Transpose means? $\endgroup$ – Error 404 Jun 14 '15 at 16:36
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    $\begingroup$ You probably want to rewrite $\sin$ in terms of $\cos$ via Pythagorean identity. Then you can turn the first equation into a quadratic equation which you can solve for $\cos\theta$. After which you can do $\arccos$ to both sides. $\endgroup$ – Cameron Williams Jun 14 '15 at 16:37
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    $\begingroup$ @VikrantDesai I think it means to "transpose the roles of the variables." Here, $\theta$ is the independent variable and $m$ is the dependent variable. OP wants to transpose their roles so that $m$ is the independent variable and $\theta$ is the dependent variable. $\endgroup$ – Cameron Williams Jun 14 '15 at 16:38
  • $\begingroup$ @CameronWilliams yeah that may be the case. $\endgroup$ – Error 404 Jun 14 '15 at 16:40
  • $\begingroup$ There are many functions that do not have an inverse in closed form. For a famous example see the Lambert W function. Do you have good reason to believe that your functions do have an expressible inverse? $\endgroup$ – Rory Daulton Jun 14 '15 at 17:21
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You may first "reduce" to $\frac{e^2\cos^2\theta-2\cos^2\theta+1}{2\cos\theta}=e-\cos\theta$

From here you derive the equation (some work left for you): $e^2\cos^2\theta-2e\cos\theta+1=0$

Solve for $\theta$ to get $\theta= arcos(1/e)$

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  • $\begingroup$ Oh, I had no idea it was a system of equations. I thought the OP meant he had to solve them individually. $\endgroup$ – John Molokach Jun 15 '15 at 9:46
  • $\begingroup$ @John, they are two independent equations $\endgroup$ – user155321 Jun 19 '15 at 6:17
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As @Cameron Williams has pointed out, you can write both of these as quadratic equations of a trigonometric function of $\theta_r.$


For equation (1), make the substitution $\cos\theta_r=\pm\sqrt{1-\sin^2\theta_r}$ and multiply through by $\sin\theta_r$ to get

$$e-m_t\sin\theta_r=\pm\sqrt{1-\sin^2\theta_r}.$$

Squaring both sides and setting equal to zero, we get

$$(m_t^2+1)\sin^2\theta_r-2em_t\sin\theta_r+(e^2-1)=0.$$

Solving for $\sin\theta_r$ using the Quadratic Formula gives

$$\sin\theta_r=\frac{em_t\pm\sqrt{m_t^2-e^2+1}}{m_t^2+1},$$

upon which you can use the inverse sine to extract $\theta_r.$


For equation (2), a similar trick will work. But as I suggested in the comments, I will use double angle formulas. Once I get to the quadratic form, I will leave it up to you to solve.

First multiplying through by $2\sin\theta_r\cos\theta_r,$ and then using the identities

  • $\sin2\theta_r=2\sin\theta_r\cos\theta_r$
  • $\cos2\theta_r=\cos^2\theta_r-\sin^2\theta_r,$ and
  • $\cos^2\theta_r=\dfrac{1+\cos2\theta_r}{2}$

we get

$$m_t\sin2\theta_r=\frac{e^2}{2}(1+\cos2\theta_r)-\cos2\theta_r.$$

Now multiplying through by 2, once again using the Pythagorean Identity $\sin2\theta_r=\pm\sqrt{1-\cos^2 2\theta_r},$ and squaring both sides, we get

$$4m_t^2(1-\cos^2 2\theta_r)=e^4+2e^2(e^2-1)\cos2\theta_r+(e^2-1)^2\cos^2 2\theta_r,$$

which is a quadratic equation in $\cos 2\theta_r$ that I will leave you to solve. smiley

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  • $\begingroup$ Hi John, thank you so much for these answers. With yours and egreg's help I'm now able to work out quadratic formulas in terms of sin, cos, tan, and their double angle counterparts. However I just want to understand Narasimham's answer before accepting an answer $\endgroup$ – user155321 Jun 21 '15 at 18:38
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Answer #2

This is if you interpret the OP's question as: "write each expression on the RHS in terms of $\tan\theta_\color{red}r,$."

For `expression' (1), $$e\csc\theta_r-\cot\theta_r=\pm e\sqrt{1+\cot\theta_r}-\cot\theta_r=\pm e\sqrt{1+\dfrac{1}{\tan\theta_r}}-\dfrac{1}{\tan\theta_r}.$$

For 'expression' (2), use the double angle formula $\cot(2\theta_r)=\dfrac{1-\tan^2\theta_r}{2\tan\theta_r}$ to get

$$\frac{e^2}{2}\cot\theta_r-\cot 2\theta_r=\frac{e^2}{2\tan\theta_r}-\frac{1-\tan^2\theta_r}{2\tan\theta_r}=\frac{e^2-1+\tan^2\theta_r}{2\tan\theta_r}.$$

Written in terms of $m_r$, expression (1) is

$$\pm e\sqrt{1+\frac{1}{m_r}}-\frac{1}{m_r},$$

and expression (2) is

$$\frac{e^2-1+m_r^2}{2m_r}.$$

Does this answer your question?

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Ah, they are separate problems-- First one is simpler. Not given.

Second one with abbreviations: Hope you guess what is omitted.

$ E = e^2/2, C = \cos 2 \theta, S =\sin 2 \theta $

$ E (1+ C)/S - C/S = m_t $

$ E + C ( E-1) = m_t S $

$ (1-E) C/A + m S / A = E/A $, where again $ A = \sqrt{(1-E)^2 + m_t^2}$

Let $ (1-E)/A = s_u , m/A = c_u $

$ s_u c _{2 \theta} + c_u s_{2 \theta} $ .... $ 2 \theta_r = \sin^{-1} (E/A) -\cos^{-1} (m_t/A) $

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  • $\begingroup$ This is exactly the same as the first answer above, where the OP noted the his/her question should not be considered a system of two equations. So setting the RHS's equal does not answer the question. $\endgroup$ – John Molokach Jun 19 '15 at 21:17
  • $\begingroup$ Thanks, not attentive, corrected it. $\endgroup$ – Narasimham Jun 19 '15 at 23:23
  • $\begingroup$ I follow your answer all the way up to the last step. I've really tried but I just can't work out the omission :( Please, is there any chance you could elaborate that last step? $\endgroup$ – user155321 Jun 21 '15 at 18:30
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The first equation is, using $m=m_t$ and $\theta=\theta_r$, $$ m\sin\theta=e-\cos\theta $$

Set $t=\tan(\theta/2)$, so you get $$ 1-t^2+2mt=e+et^2 $$ that becomes the standard quadratic $$ (1+e)t^2-2mt+e-1=0 $$ Just observe that $\theta=\pi$ is not a solution unless $e=-1$, case that you can treat separately.

The second equation can be written $$ \sin^2\theta-2m\sin\theta\cos\theta+(e^2-1)\cos^2\theta=0 $$ Note that $\cos\theta\ne0$, so this becomes $$ \tan^2\theta-2m\tan\theta+e^2-1=0 $$ which is a quadratic in $\tan\theta$.

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  • $\begingroup$ Hi egreg, thank you so much for these answers. With yours and John's help I'm now able to work out quadratic formulas in terms of sin, cos, tan, and their double angle counterparts. However I just want to understand Narasimham's answer before accepting an answer $\endgroup$ – user155321 Jun 21 '15 at 18:43

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