1. ## Trigonometry

How to solve the following:

-10t + 60/piSine(pi/12 * t) + 2400

2. isn't it -5t

3. Nope. -10t

4. Originally Posted by Nanpa
How to solve the following:

-10t + 60/piSine(pi/12 * t) + 2400
This is an expression and cannot be solved. It would have to be an
equation to have a solution.

The best you can do with an expression is simplify it.

Also the expression is ambiguous as it stands, what you have written
could be intended to mean:

$\displaystyle -10t + \frac{60}{\pi} \sin(\pi.t / 12 ) + 2400$

or:

$\displaystyle -10t + \frac{60}{\pi \sin(\pi.t / 12 )} + 2400$

or:

$\displaystyle -10t + \frac{60}{\pi} sin(\pi / (12t )) + 2400$,

or:

$\displaystyle -10t + \frac{60}{\pi \sin(\pi./ (12t) )} + 2400$

Now you probably intend the first of these, but you should use brackets
to make the meaning unambiguous when typing maths in ASCII.

RonL

5. Sorry, I forgot to add equals zero...

6. he means the top one captain

7. t=240

8. Thanks, but how?

9. How you work that out???

10. Originally Posted by nath_quam
How you work that out???
$\displaystyle -10t + \frac{60}{\pi} \sin(\pi.t / 12 ) + 2400=0$

First look for an approximate solution. Since the sinusoidal term is always
between $\displaystyle \pm 60/pi \approx \pm20$, it is clear that a solution
is close to a solution of $\displaystyle -10t+2400=0$, which occurs when
$\displaystyle t=240$, now the $\displaystyle \sin$ term is identically zero
at that point, so this is an exact root.

That there are no other roots is obvious as the maximum rate of increase
of the middle term is less than $\displaystyle 10$ (in fact it is $\displaystyle 5$)
so the entire LHS is a decreasing function and so the root is unique.

RonL