For with where , find in terms of .

I am terrible at maths so please show any simple easy to understand working out so that I can understand better.

Any help would be appreciated!

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- April 30th 2011, 10:51 AMJoker37Find a in terms of b
For with where , find in terms of .

I am terrible at maths so please show any simple easy to understand working out so that I can understand better.

Any help would be appreciated! - April 30th 2011, 12:16 PMHappyJoe
Recall the following "transition formula" for going from cosine to sine, where is arbitrary:

Since the numbers and are different by , and since sine is periodic with period , we have also that

whence

Since , we have that lies in the interval

Since you want to solve for , we have by the transition formula above that we might just as well solve

for , where the last equality follows from sine being an odd function.

Since , we have that lies in the interval . Hence both and are numbers in the interval , on which sine is injective, so in solving

for , we can apply to both sides to obtain

or

- April 30th 2011, 07:25 PMJoker37
I don't understand all of the above.

Below was my attempt:

However the answer is:

Where did I go wrong? - April 30th 2011, 07:29 PMtopsquark
- April 30th 2011, 07:31 PMJoker37
- April 30th 2011, 08:01 PMtopsquark
- April 30th 2011, 08:13 PMJoker37
- April 30th 2011, 08:38 PMtopsquark