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- Sep 23rd 2007, 04:48 PMqbkr212 Math Question Involving Inverse Trig
Thanks!!:D

- Sep 23rd 2007, 05:04 PMJhevon
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 - Sep 23rd 2007, 06:26 PMThePerfectHacker
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 . - Sep 23rd 2007, 06:32 PMKrizalid
Or we can take

, and evaluate

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For the second problem, find the sine in terms of the tangent, the apply the same method as above. - Sep 23rd 2007, 06:46 PMThePerfectHacker
This equation is not correct!

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, . - Sep 23rd 2007, 07:02 PMqbkr21Re:
- Sep 23rd 2007, 07:15 PMThePerfectHacker
The hypothenuse is the side is call the other side then by Pythagorus, so .

- Sep 23rd 2007, 07:16 PMJhevon
- Sep 23rd 2007, 07:23 PMqbkr21Re:
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

- Sep 23rd 2007, 07:31 PMJhevon
- Sep 23rd 2007, 07:32 PMqbkr21Re:
Sorry...I was thinking of Problem #2. Let me get to work on it so I can see if I have any questions.

Thanks,

-qbkr21 - Sep 24th 2007, 08:04 AMKrizalid
- Sep 24th 2007, 10:08 AMThePerfectHacker