# How do you isolate $\theta$ from $\sin(\theta) - k\cos(\theta) = \frac{m_1}{m_2}$?

This equation derives from physics force equations, and I have verified to make sure that the equation works. No actual physics pertains to this question. The part I need help on only requires math skills. Here is the equation:

$$\sin(\theta) - k\cos(\theta) = \frac{m_1}{m_2}$$

where $$k$$, $$m_1$$, and $$m_2$$ are variables, specifically the kinetic friction coefficient, mass one, and mass two respectively (irrelevant).

I have tried squaring both sides to then use double angle rules on the $$2\sin\cos$$ that yields, but all of my attempts have failed to simplify the equation further.

Could someone help me derive an equation for theta?

Let $$sin{\theta} = x$$.

$$x-k\sqrt{1-x^{2}} = \frac{m1}{m2}$$

Separate the linear terms and square both sides. $$(x-\frac{m1}{m2})^{2} = k^{2}(1-x^{2})$$

It is now a simple quadratic equation which you can solve. But remember to choose only those solutions that lie within [-1,1] or can lie within [-1,1] for some values of $$k,m1, m2$$.

Now when you have obtained $$\sin{\theta}$$, use $$\sin^{-1}$$ to obtain $$\theta$$

It can be shown that any expression of the form $$a\cos(\theta)+b\sin(\theta)$$ ($$a$$ and $$b$$ are constants) can be written in the form $$m\cos(\theta+c)$$, where $$m$$ and $$c$$ are constants. Specifically, by taking $$m=\text{sgn}(a)\sqrt{a^2+b^2}$$ and $$c=-\tan^{-1}\left(\frac{b}{a}\right)$$, where $$\text{sgn}(a)$$ denotes the sign of $$a$$, we can write

$$a\cos(\theta)+b\sin(\theta)=\text{sgn}(a)\sqrt{a^2+b^2}\cos\left(\theta-\tan^{-1}\frac{b}{a}\right)$$

Substituting $$a=-k$$ and $$b=1$$ gives

\begin{align*} \sin(\theta)-k\cos(\theta) &= \text{sgn}(-k)\sqrt{(-k)^2+1^2}\cos\left(\theta-\tan^{-1}\frac{1}{-k}\right)\\ &=-\text{sgn}(k)\sqrt{1+k^2}\cos\left(\theta+\tan^{-1}\frac{1}{k}\right) \end{align*}

so if $$k>0$$, your equation boils down to

$$-\sqrt{1+k^2}\cos\left(\theta+\cot^{-1}k\right)=\frac{m_1}{m_2}$$

Dividing both sides by $$-\sqrt{1+k^2}$$ gives $$\cos\left(\theta+\cot^{-1}k\right)=-\frac{m_1}{m_2\sqrt{1+k^2}}$$, provided that $$\left|-\frac{m_1}{m_2\sqrt{1+k^2}}\right|\leq 1$$.

It can be shown that for any real number $$y$$ with $$|y|\leq 1$$, the only solutions to the equation $$\cos(x)=y$$ are of the form $$x=2\pi m\pm\arccos(y)$$ for integer $$m$$, so

$$\theta+\cot^{-1}(k)=2\pi m\pm\arccos\left(-\frac{m_1}{m_2\sqrt{1+k^2}}\right)$$

The desired result immediately follows.

$$\theta=2\pi m-\cot^{-1}(k)\pm\arccos\left(-\frac{m_1}{m_2\sqrt{1+k^2}}\right)$$

You might want to post the physics question which led you to this equation. It is likely there was a mistake in the derivation. Nonetheless,

$$\sin \theta - k \cos \theta = m_1/m_2 = \gamma$$ $$\sin \theta - k\sqrt{1-\sin^2 \theta} = \gamma$$ $$\sin \theta - \gamma = k \sqrt{1-\sin^2 \theta}$$ $$(\sin \theta -\gamma)^2 = k^2(1-\sin^2 \theta)$$ $$\sin^2 \theta - 2\sin \theta \gamma + \gamma^2 = k^2-k^2 \sin^2 \theta$$ $$(k^2+1) \sin^2 \theta +(-2 \gamma) \sin \theta + \gamma^2 -k^2=0$$ $$\sin \theta = \frac{2\gamma \pm \sqrt{4\gamma^2 - 4(k^2+1)(\gamma^2 -k^2)}}{2(k^2+1)}$$

• I believe this is a small mistake, but shouldn't "k" be squared in the fourth step? Nov 19, 2020 at 6:02
• @Quadvortex Yes, I've fixed that now Nov 19, 2020 at 6:05

Let:

$$k=\tan \alpha \ \iff \ \alpha:=\tan^{-1}k.$$

The equation

$$\sin \theta - k \cos \theta = \underbrace{m_1/m_2}_{\gamma}$$

becomes, by multiplication by $$\cos \alpha$$:

$$\cos \alpha \sin \theta - \sin \alpha \cos \theta =\gamma \cos \alpha$$

$$\sin(\theta - \alpha)=\gamma \cos \alpha \tag{1}$$

If $$\gamma \cos \alpha \in [-1,1]$$, we can set

$$\sin \delta:= \gamma \cos \alpha \ \iff \ \delta=\sin^{-1}(\gamma \cos \alpha)$$

Equation (1) becomes:

$$\sin(\theta - \alpha)=\sin \delta \tag{2}$$

$$\iff \begin{cases}\theta - \alpha&=&\delta+k 2 \pi\\ \theta - \alpha&=&\pi-\delta+k' 2 \pi\end{cases}$$

whence two expressions for $$\theta$$.

Here's a general guide and explanation for problems of your type:

If we have an expression, $$a\sin{x}+b\cos{x}$$, let us assume it can be written in the form $$R\sin(x+\alpha)$$ Now to see if we can find values for $$R$$ and $$\alpha$$ in terms of $$a$$ and $$b$$. Using the compund angle formulae, also known as the addition formulae: $$R\sin(x+\alpha)=R\sin{x}\cos{\alpha}+R\sin\alpha\cos x=a\sin{x}+b\cos{x}$$ So we have $$R\cos\alpha=a,R\sin\alpha=b$$ So dividing the second equality by the first: $$\tan\alpha=\frac{b}{a}$$ meaning we can find $$\alpha$$ in terms of $$a$$ and $$b$$, as we wanted. Now, to find $$R$$: Squaring the $$2$$ equalities above we have $$R^2\cos^2\alpha+R^2\sin^2\alpha=R^2(\cos^2\alpha+\sin^2\alpha)=R^2=a^2+b^2\implies R=\sqrt{a^2+b^2}$$ So, to finish off by recapping what we have learnt: $$\tan\alpha=\frac{b}{a},R=\sqrt{a^2+b^2}$$ meaning $$a\sin{x}+b\cos{x}\equiv\sqrt{a^2+b^2}\sin(\theta+\arctan{\frac{b}{a}})$$ Try applying that to your question. If you have any questions, please ask!

I hope that was helpful :)

Note: It's not always going to be the case that we take the positive square root for $$R$$. It depends on the sign of the quantities $$a$$ and $$\cos\alpha$$ ( I could've picked $$b$$ and $$\sin\alpha$$ as well; can you see why?).