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I want to use the idea of vectors to solve this problem: "How many times a day do the hour and minute hands of a clock face the opposite of each other?" We know that the minute hand turns $6^{\circ}$ every minute. If I assume my starting point is 12 o'clock, then my angle (with respect to the positive x-axis, of course), $\theta_m(t)=90^{\circ}-6t$. Similarly, the angle between the hour-hand and the positive x-axis at any given time (in minutes), t, is $\theta_h= 90^{\circ}- \frac{1}{2}t$. If we think of the hour and minute hands as two dimensional vectors whose tails are on the origin, then the vector of the minute hand can be specified as $\textbf{m}$, where: \begin{align*} \textbf{m} &= \begin{bmatrix} \sin(6t) \\ \cos(6t) \\ \end{bmatrix} \end{align*} In the same way, the vector of the hour hand, $\textbf{h}$= \begin{bmatrix} \sin( \frac{1}{2}t) \\ \cos( \frac{1}{2}t) \\ \end{bmatrix}

We want the instances where: \begin{align*} \textbf{m} &= -\textbf{h} \\ \textbf{m}+\textbf{h} &=\textbf{0} \\ \begin{bmatrix} \sin(6t)+\sin\left( \frac{1}{2}t\right) \\ \cos(6t)+\cos\left( \frac{1}{2}t\right) \\ \end{bmatrix} &= \textbf{0} \end{align*} So we have: $$\sin(6t)+\sin\left( \frac{1}{2}t\right) -\cos(6t)-\cos\left( \frac{1}{2}t\right)=0$$ In radians: $$\sin\left( \frac{\pi t}{30}\right)+\sin\left( \frac{\pi t}{360}\right) -\cos\left( \frac{\pi t}{30}\right)-\cos\left( \frac{\pi t}{360}\right)=0$$ My question is, can I find out the roots of that by hand (for my puposes, $0<t \le 1440$)? I plotted it and the period seems to be changing.

Edit: So technically from the vector equation I get this: $$sin(\frac{\pi t}{30})+sin(\frac{\pi t}{360})=0$$ $$cos(\frac{\pi t}{30})+cos(\frac{\pi t}{360})=0$$

It turns solving these get me the true answers, while solving the one in the title gives me unnecessary ones (I'm saying all of this by looking at the plots). So, how do I solve these 2 sets by hand?

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Hint:

Let $\dfrac{\pi x}{360}=t$

$$\sin12t+\sin t=\cos12t+\cos t$$

Use Prosthaphaeresis Formulas to find $$2\sin\dfrac{13t}2\cos\dfrac{11t}2=2\cos\dfrac{13t}2\cos\dfrac{11t}2$$

So, either $\cos\dfrac{11t}2=0\implies\dfrac{11t}2=(2n+1)\dfrac\pi2$ where $n$ is any integer

or $\sin\dfrac{13t}2=\cos\dfrac{13t}2\iff\tan\dfrac{13t}2=1\implies\dfrac{13t}2=m\pi+\dfrac\pi4$ where $m$ is any integer

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  • $\begingroup$ @cosmo5, Any doubt :) Thanks $\endgroup$ Jan 5, 2021 at 13:40
  • $\begingroup$ I edited my post. Could you take a look at it? $\endgroup$ Jan 5, 2021 at 16:30
  • $\begingroup$ You made a mistake: you substituted y for t. $\endgroup$ Jan 5, 2021 at 16:43
  • $\begingroup$ @OrlinAurum, Is it fine, now? $\endgroup$ Jan 5, 2021 at 17:45
  • $\begingroup$ Yes. I just want to make it clear for future readers that the t in your answer is not the same as the t in my post. $\endgroup$ Jan 5, 2021 at 17:58
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Hint:

Method$\#1:$ $$\sin6t-\cos6t=\cos\dfrac t2-\sin\dfrac t2$$

Using Solving trigonometric equations of the form $a\sin x + b\cos x = c$ or Solving $a\sin x + b\cos x = c $

$$\sin\left(6t-\dfrac\pi4\right)=\sin\left(\dfrac\pi4-\dfrac t2\right)$$

OR Method$\#2:$

$$\cos6t-\sin6t=-\left(\cos\dfrac t2-\sin\dfrac t2\right)$$

$$\implies\cos\left(6t+\dfrac\pi4\right)=-\cos\left(\dfrac t2+\dfrac\pi4\right)$$

Again, $$-\cos\left(\dfrac t2+\dfrac\pi4\right)=\cos\left(\dfrac t2+\dfrac\pi4-\pi\right)$$

Can you take it from here?

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Ok, we have $$sin( \frac{\pi t}{30})+sin( \frac{\pi t}{360})=0$$ Let $ \frac{\pi t}{360}=y$. So we have: $$sin(12y)+sin(y)=0$$ As lan bhattacharjee pointed out, we could use Simpson's formulas here: \begin{align*} sin(12y)+sin(y) &= 2sin[ \frac{1}{2}(12y+y)]cos[ \frac{1}{2}(12y-y)] \\ &= 2sin( \frac{13y}{2})cos( \frac{11y}{2}) \end{align*} So we have: \begin{equation} 2sin( \frac{13y}{2})cos( \frac{11y}{2})=0 \end{equation}

From the main post it is also given: $$cos( \frac{\pi t}{30})+cos( \frac{\pi t}{360})=0$$ From this we get: \begin{equation} 2cos( \frac{13y}{2})cos( \frac{11y}{2})=0 \end{equation}

Comparing (1) & (2), it is apparent that $cos( \frac{11y}{2})$ has to be zero, because if it isn't zero, then we get $sin( \frac{13y}{2})=cos( \frac{13}{y})$. But we know that $sinx=cosx$ only when $sinx=cosx= \frac{1}{\sqrt{2}}$. Yet, we assumed that $cos( \frac{11y}{2})$ doesn't equal zero. That would leave us with $sin( \frac{13y}{2})cos( \frac{11y}{2}) \neq 0$, which is not true. So the only way for equations 1 and 2 to be true is if: \begin{align*} cos( \frac{11y}{2}) &= 0 \\ \frac{11y}{2} &= \frac{\pi}{2}(2n+1), n \in \mathbb{Z} \\ y &= \frac{\pi}{11}(2n+1) \\ \frac{\pi t}{360} &= \frac{\pi}{11}(2n+1) \\ t &= \frac{360}{11}(2n+1) \end{align*}

Since we're dealing with time, we want: \begin{align*} t &> 0 \\ ... &> ... \\ n &\geq 0 \end{align*} There are 1440 minutes in a single day: \begin{align*} t &\leq 1440 \\ ... &\leq ... \\ n &\leq 21 \end{align*}

There are 22 integers between 0 and 22 (including 0 and 22). So, the hour and minute hands of a clock face opposite positions to each other 22 times a day.

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