sin(t)=a
t=(-1)^k sin^(-1) (a) + πk
please explain I do not understand the (-1)^k and πk
Since sine has a period of 2π shouldnt the equation for all solution be t=sin^(-1)(a) +2πk?
where does the (-1)^k and πk come from. How is this derived?
sin(t)=a
t=(-1)^k sin^(-1) (a) + πk
please explain I do not understand the (-1)^k and πk
Since sine has a period of 2π shouldnt the equation for all solution be t=sin^(-1)(a) +2πk?
where does the (-1)^k and πk come from. How is this derived?
Frankly I have no idea what you meant in the above posting.
To say that $\sin(\theta)=a$ means that $-1\le a\le 1$ but that says nothing about $\large\bf{\theta}$.
If $\frac{-\pi}{2}\le\theta\le\frac{\pi}{2}$ then $\sin(\theta)$ takes on every possible value in $[-1,1]$.
However, because these functions are periodic then $\sin(\theta)=\sin(\theta\pm 2k\pi),~k\in\mathbb{Z}$
Now let us discuss the basics. The function $\arcsin(t)$ maps the interval $[-1,1]\to \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]$
Now look at this webpage.
While the principal value is $\dfrac{-\pi}{4}$ there are infinitely many other solutions: $\dfrac{-\pi}{4}+2k\pi,~k\in\mathbb{Z}$
@Melody2
It may have been that I pointed out a more compact way of doing it.
In any case PLEASE DO NOT DELETE YOUR POST. It may help the student more than I can.
It has been at least eleven years since I have taught an undergraduate course.
So you may well be much be better than I at helping this student.