Sin(pi/2) (different from sin )

[phymatter]

sin(pi/2) (different from sin!)

how can sin(pi/2) be equal to 1 , i mean sinx is defined as the ratio P/H of a right triangle , but if sinx = 1
=> P=H , but by Phytogoras
=> B=0 ,
but how is this possible ?
so how can sinx =1 exist?

 

[arildno]

how can sin(pi/2) be equal to 1 , i mean sinx is defined as the ratio P/H of a right triangle , but if sinx = 1
=> P=H , but by Phytogoras
=> B=0 ,
but how is this possible ?
so how can sinx =1 exist?

A good question!

Answer:
The sine function to which sin(pi/2) is given meaning, is DEFINED DIFFERENTLY from regarding it as the ratio between the opposite side/hypotenuse of a right-angled triangle.

 

[phymatter]

The sine function to which sin(pi/2) is given meaning, is DEFINED DIFFERENTLY from regarding it as the ratio between the opposite side/hypotenuse of a right-angled triangle.

What's that defination ? please tell me that to me!

 

[Hootenanny]

how can sin(pi/2) be equal to 1 , i mean sinx is defined as the ratio P/H of a right triangle , but if sinx = 1
=> P=H , but by Phytogoras
=> B=0 ,
but how is this possible ?
so how can sinx =1 exist?

A question for you: How can the angle between the hypotenuses and any side of a triangle ever be pi/2?

 

[arildno]

What's that defination ? please tell me that to me!

Sure!

1. Draw a unit circle in the (x,y)-plane with the origin as its centre.

2. Now, you can think in terms of angles between the positive x half-axis, and any other ray in the plane, emanating from the origin.

3. The sine value to a particular angle is defined to equal the y-coordinate of the point of intersection the associated ray makes with the circle.
EXAMPLE:
4. If you let the full circle be given the angular value $2\pi$, then the angle the positive "y"-half axis/ray makes with the positiv x half-axis will be 1/4th of that, namely [itex]\frac{\pi}{2}[/tex]<br /> <br /> 5. The point of intersection between that ray (the positive "y" half-axis) makes with the unit circle is (0,1), whose "y"-coordinate is..1.<br /> Therefore, in this context, $\sin(\frac{\pi}{2})=1$[/itex]

 

[HallsofIvy]

First, the answer to your original question: since there cannot exist a right triangle with two right angles, when we talk about sin(pi/2) we are actually taking a limit. As x approaches pi/2, sin(x) becomes arbitrarily close to pi/2.

Now, your second question:
There are a number of equivalent definitions of sine and cosine independent of right triangles.

1) Draw the unit circle on a coordinate system. For non-negative t, starting at point (1, 0), measure counter-clockwise around the circumference of the circle. The point at distance t, around the circumference, from (1, 0) has coordinates (cos(t), sin(t)), by definition. Since the unit circle has circumference pi, pi/2 is 1/4 of the way around the circle, pi/2 takes us 1/4 of the way around the circumference to (0, 1). cos(pi/2)= 0, sin(pi/2)= 1. Notice that the x and y coordinates, and so cosine and sine can have negative values also. Going a distance "pi" around the circle takes us half way around, to the point (-1, 0): cos(pi)= -1, sin(pi)= 0.

That is the simplest way to define sin(t) and cos(t) but there are some technical difficulties with mathematically defining "measuring along the circumference". To avoid that, you could also:

2) Define sin(x) to be the power series
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1}$$
and define cos(x) to be the power series
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$$

Or, my personal preference,
3) Define sine and cosine to be the solutions to given "initial value problems":

y(x)= sin(x) is the function satisfying the differential equation y"= -y, with initial values y(0)= 0, y'(0)= 1.

y(x)= cos(x) is the function satisfying the differential equation y"= -y, with initial values y(0)= 1, y'(0)= 0.

Another advantage of those definitions is that you can prove the most important property of sine and cosine, that they are "periodic" functions, which does not even make sense for the "right triangle" definitions where sine and cosine are not defined for all t.

 

[arildno]

phymatter:

Note that the "strange", or "overly mathematical" definitions of HallsofIvy are motivated by some rather subtle arguments about what we mean by "measurement".

Therefore, mathematicians have chosen to hone the definitions to get around those problems, and get definitions that are actually better, within mathematical theory.

However, as Halls agrees with me, I am sure, the best pedagogical method is to dispense with the subtle measurement problem, and instead give an (apparently) simple GEOMETRIC interpretation of the sine function.

 

[phymatter]

Thank you Everyone!