• Mar 4th 2011, 03:10 AM
cupid
suppose i have a function sin(sin x)
sin x will give values [-1,1] ...
so now, should i take these values in degree or radian to find sin(sin x) ?
• Mar 4th 2011, 03:41 AM
FernandoRevilla
The standard "geometric" definition for:

$\displaystyle \sin:\mathbb{R}\to [-1,1],\quad x\to \sin x$

$\displaystyle \cos: \mathbb{R}\to [-1,1],\quad x\to \cos x$

is when $\displaystyle x$ is "measured" in radians. So we have good properties, for example:

(i) $\displaystyle (\sin)'=\cos$

(ii) $\displaystyle \sin x=x-x^3/3!+x^5/5!-\ldots\quad (\forall x\in \mathbb{R})$

etc.
• Mar 4th 2011, 11:14 AM
Perfessor
• Mar 5th 2011, 01:19 AM
HallsofIvy
The only time you should use degrees is when the problem specifically gives angles in degrees. For problems in which sine (or cosine) is used just as a function always use radians.

(There are a number of ways to define sine and cosine that do not involve angles at all. So strictly speaking the argument is not in "radians" or "degrees". However, to be able to identify those functions with the sine and cosine functions we learned in trigonometry, we use radians.)
• Mar 6th 2011, 04:12 AM
cupid
Quote:

Originally Posted by HallsofIvy
The only time you should use degrees is when the problem specifically gives angles in degrees. For problems in which sine (or cosine) is used just as a function always use radians.

(There are a number of ways to define sine and cosine that do not involve angles at all. So strictly speaking the argument is not in "radians" or "degrees". However, to be able to identify those functions with the sine and cosine functions we learned in trigonometry, we use radians.)

Actually My question is that the output of sine and cosine functions is in degrees or radians
• Mar 6th 2011, 04:29 AM
Plato
Quote:

Originally Posted by cupid
Actually My question is that the output of sine and cosine functions is in degrees or radians

For every $\displaystyle x\in\mathbb{R}$ it is the case that $\displaystyle \sin(x)\in\mathbb{R}$ and $\displaystyle \cos(x)\in\mathbb{R}$.
• Mar 6th 2011, 07:56 AM
cupid
Quote:

Originally Posted by Plato
For every $\displaystyle x\in\mathbb{R}$ it is the case that $\displaystyle \sin(x)\in\mathbb{R}$ and $\displaystyle \cos(x)\in\mathbb{R}$.

Isn't it more like ...

$\displaystyle sinx \in [-1,1]$ ???
• Mar 6th 2011, 08:02 AM
Plato
Quote:

Originally Posted by cupid
Isn't it more like ...$\displaystyle sinx \in [-1,1]$ ???

But $\displaystyle [-1,1]\subset\mathbb{R}$.

The point being that both the sine and cosine functions map real numbers to real numbers.
• Mar 6th 2011, 08:10 AM
cupid
but my question is that ... for example i have a question ... sin (sin x)

i need to calculate the value of this function at $\displaystyle x = \frac{\pi}{2}$

as $\displaystyle sin\frac{\pi}{2} = 1$
now i need to calculate sin(1)

so should i find value of sin(1 radian) or sin(1 degree) ???
• Mar 6th 2011, 08:19 AM
Plato
Quote:

Originally Posted by cupid
but my question is that ... for example i have a question ... sin (sin x)
i need to calculate the value of this function at $\displaystyle x = \frac{\pi}{2}$
as $\displaystyle sin\frac{\pi}{2} = 1$ now i need to calculate sin(1)
so should i find value of sin(1 radian) or sin(1 degree) ???

I do not know what a degree is.
I know what real numbers are.
1 is a real number.
Thus it is $\displaystyle \sin\left(\sin\left(\frac{\pi}{2}\right)\right)=\s in(1)\approx 0.841$.
• Mar 6th 2011, 11:29 AM
cupid
Quote:

Originally Posted by Plato
I do not know what a degree is.
I know what real numbers are.
1 is a real number.
Thus it is $\displaystyle \sin\left(\sin\left(\frac{\pi}{2}\right)\right)=\s in(1)\approx 0.841$.

that is for 1 radian ... for 1 degree its like 0.0014 something
• Mar 6th 2011, 11:48 AM
TheEmptySet
The point that I think you are missing is that Radians are a pure number! They are defined as the arc length of a circle divided by the radius of the same circle so strictly speaking the "units" (length) reduce out leaving a Real number.

Radian - Wikipedia, the free encyclopedia
• Mar 6th 2011, 12:14 PM
HallsofIvy
Yes, that was what I was trying to say to begin with!

"Strictly speaking the argument is not in "radians" or "degrees". However, to be able to identify those functions with the sine and cosine functions we learned in trigonometry, we use radians."