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 .
Or we can take
, and evaluate
For the second problem, find the sine in terms of the tangent, the apply the same method as above.
What if . (Fubar)
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, .
The hypothenuse is the side is call the other side then by Pythagorus, so .
One last thing. If we are looking for cos(Ө) that would be adjacent/hypotenuse. However this wouldn't the solution I am looking for. Adjacent/Hypotenuse would instead give me
Sorry...I was thinking of Problem #2. Let me get to work on it so I can see if I have any questions.