Sinus domain: y = sin(alpha) What is the domain of y is between -1 to 1?

[klimbo]

Let's just talk now about why sinx can be between 0 and 1. Consider a right angle with the base leg having a length of 1 (any other number will work). Let the angle between the base leg and the hypotenuse be x. Lets call the opposite side y. If y has length of 0, then the angle x = 0^o. So the sin(0) = opp/hyp = 0/1 = 0. I say that the hypotenuse is 1 since if a right angle has a 0^o angle, then the hypotenuse lays right on top of the base which has length = 1). Now consider what happens if y is extremely large (and the base is still 1). What happens is the opp and hyp are about the same length. In fact if we keep letting y get larger and other, then sin(x) gets closer and closer to 1. In fact the sin (90^o) =1. Since the sin of some angles can be negative, we also have that the sin(x) can be between -1 and 0. Pnr Status TextNow VPN Combining we have sin(x) is between -1 and 1

exactly i agreed qwith you the sin(x) can be between -1 and 0 but with fort probabillity to be 1

 

[Otis]

because the sin of an angle is never larger than … -1 …

You're thinking in terms of absolute values, but we can't express the range like that.

… the sin of an angle is always greater than or equal to -1

I think that explanation could be better worded, too.


\(\displaystyle |sin(\theta)| \le 1\)

or

\(\displaystyle -1 \le sin(\theta) \le 1\)

:cool:

 

[pka]

You're thinking in terms of absolute values, but we can't express the range like that.
I think that explanation could be better worded, too.
\(\displaystyle sin(\theta) \le |1|\)
or
\(\displaystyle -1 \le sin(\theta) \le 1\)

\(\displaystyle |1|=1\) therefore if \(\displaystyle f(x)\le|1|\) then \(\displaystyle f(x)\le 1\) period.

I think you mean if \(\displaystyle |f(x)|\le 1\) then \(\displaystyle -1\le f(x)\le 1\) do you not?

 

[Otis]

… I think you mean if \(\displaystyle |f(x)|\le 1\) …

Good grief, yes. (Must be going senile.)