In order to see if there's some kind of strategy with this, I'm trying the desired identity with , giving
How would you go about showing this sort of thing?
For example, is there a way of writing in terms of functions with an argument of ?
Okay, I'm working on a really tricky induction problem where I have to show the following (assuming that ):
for all natural numbers .
I've already verified the base case (with the help of ebaines), so now I'm trying to show that the above equation implies this equation:
I almost hesitate to ask this as to me it seems either impossible or a huge amount of work, but perhaps that's not so for those with more experience. Using the first equation we can write the second equation as
so what we desire to show is that
So I'm unsure where to go from here. We already have the denominators set up neatly, so I don't want to mess with that. I thought about distributing the in the third term and then adding the resulting two terms to the first two terms in some manner, but I'm not sure how the arguments will work out. Can anyone help?
In order to see if there's some kind of strategy with this, I'm trying the desired identity with , giving
How would you go about showing this sort of thing?
For example, is there a way of writing in terms of functions with an argument of ?
Nice! I am far too lazy to type up Chegg's whole solution but the main steps were (starting from the LHS of the desired identity, last one in my first post):
Got a common denominator.
Rewrote in the numerator as
(half-angle formula).
Distributed .
Added terms.
Factored out of remaining terms.
Used product identies on and terms.
Simplified.
Used a difference identity on .
Simplified and divided by denominator.
Okay, I used the same identities but attacked it in a different way, (having failed to find my way through the induction approach).
Start by letting
Then, multiplying by
Now subtract the original series,
So,
or, multiplying both sides by a negative sign,
So, finally
Hope there aren't any misprints !