# trigonometry equations Take this question:

"We follow the tips of the hands of an old fashioned analog clock (360 degrees is 12 hours) . We take the clock and put it into an axis system. The origin (0,0) of the axis system is the rotationpoint of the hands. The positive x-as goes through "3 hour" and the positive y-as through "12 hour" We calculate the time "t" in hours, starting from 0:00 hours.

The equation for the tip of the big clockhand is: x=3sin2πt, y=3cost2πt

The equation for the tip of the small clockhand is x=2sin(1/6)πt, y=2cos(1/6)πt

On t=0 the two hands overlap eachother. Calculate the first point in time after t=0 when this occurs.

SOLUTION:

"This is true when cos(2πt)=cost(1/6πt) and sin(2πt)=sin(1/6πt) So, t = 12/11"

I simply don't know where to start...

• I think you want $x=3 \cos{2 \pi t}$... – Ron Gordon Mar 24 '13 at 13:32
• please elaborate. – user1095332 Mar 24 '13 at 13:35
• In fact, as I read through this problem, the whole thing makes no sense whatsoever. How on earth does one make an equilateral triangle from a side of $2$ and another of $3$? As for the cosine, read your statement as to the positive $x$ axis. – Ron Gordon Mar 24 '13 at 13:41
• different question above – user1095332 Mar 24 '13 at 13:55

Let our time $t$ be $x$ hours and $y$ minutes where $0 \leq x < 12$ and $0 \leq y < 60$.

Let $\theta_1$ be the angle of the minute hand and $\theta_2$ be the angle of the hour hand where we measure angles clockwise beginning at 12 o'clock so that $0 \leq \theta_1, \theta_2 < 360^{\circ}$

It follows that $\theta_1 = 360 \times \frac{y}{60}$ and $\theta_2 = 360 \times \frac{x}{12} + 360\times \frac{y}{60\times 12} = 360(\frac{x}{12} + \frac{y}{60\times 12})$.

If the angle of the hour hand is the same as the minute hand, we have $\theta_1 = \theta_2$ and hence:

$$\frac{y}{60} = \frac{x}{12} + \frac{y}{60\times 12}$$

We know that the hour and minute hands will have equal angles at some time shortly after 1 o'clock, so setting $x=1$ gives:

$$\frac{y}{60} - \frac{y}{60 \times 12} = \frac{1}{12} \iff \frac{12y-y}{60\times 12} = \frac{1}{12}$$

Solving for $y$:

$$11y = 60 \iff y = 60/11$$

So the two angles are equal $1$ hour and $60/11$ minutes after 12 o'clock, which is $1 + 1/11 = 12/11$ hours after 12 o'clock.

• Thanks, but what about the formulas? How to use them in my calculator? – user1095332 Mar 24 '13 at 18:18

For the overlap thing, the next time there's an over lap is when the minute hand has already gone through a full hour (revolution). Then

$$2 \pi (t+1) = \frac{\pi}{6} t \implies t + 1 = \frac{12}{11}$$

• How did you get: $$2 \pi (t+1) = \frac{\pi}{6} t$$? – user1095332 Mar 24 '13 at 14:18
• $t$ is in hours, as you said. As I said, the next time there's an overlap is when the minute hand goes through a full revolution. Thus, the $t+1$. – Ron Gordon Mar 24 '13 at 14:19
• $$t + 1 = \frac{12}{11}$$ Does that mean $$t = \frac{12}{11} - 1$$ – user1095332 Mar 24 '13 at 14:20
• It does not seem to work when I plot it on wolframalpha nor my TI-84 What should I plot on my TI calculator? – user1095332 Mar 24 '13 at 14:22