Solving for $\theta$ in trigonometric equations

Question

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,

Answer 1

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 $\theta= arcos(1/e)$

Answer 2

As @Cameron Williams has pointed out, you can write both of these as quadratic equations of a trigonometric function of $\theta_r.$

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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 $\theta_r.$

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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.

Answer 3

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?

Answer 4

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) $

Answer 5

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 $\tan\theta$.