Thanks!!
Let
so now we can construct a right triangle with an acute angle , where the opposite side to that angle is and the hypotenuse is 1. thus, by Pythagoras' theorem, we can find the adjacent side, which will be
Now, finally, we can say, ?? that final piece is for you
you can try a similar method for the second
To Show: .
Do the following steps in order.
1)Define the function .
2)Show that is in the domain of this function.
3)Show that is continous on .
4)Show that is differenciable on .
5)Show that on .
6)Notice that on
7)Therefore argue that on .
8)Prove that .
9)Therefore, on .
10)But and .
11)Finally conclude that on .
This equation is not correct!
What if .
The correct equation is much more difficult.
.
Where is the so-called "Krizalid function" it is defined as follows:
and , i.e. the function is periodic.
So for example, we get it we add we get back is also . If we subtract we get back it is here also. But is not among any of these periodic intervals. So it means , thus, .