Difficult Trigonometric Proofs

I know I'm double posting but... anyone?

1. Start with the left hand side LHS:

8cos^4(O)

Note this can be re-written as
8[cos^2(O)]^2..........................(1)

And we know the identity:
cos^2(O) = [1+cos(2O)]/2................(2)

Square both sides:
[cos^2(O)]^2 = [1+2cos(2O)+(cos(2O))^2]/4.................(3)

Use identity (2) again on the RHS of (3):
[cos^2(O)]^2 = [1+2cos(2O)+[1+cos(4O)]/2 ]/4..........(4)

Can you see where (4) comes from? Instead of O in (2), we use 2O.

[cos^2(O)]^2 = [1.5+2cos(2O)+0.5cos(4O)]/4

Multiply both sides by 8:
8[cos^2(O)]^2 = 2[1.5+2cos(2O)+0.5cos(4O)]
= cos(4O) + 4cos(2O) + 3

Thus LHS = RHS and the identity is proven.
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2. 2cos6(O)=64cos^6(O) - 96cos^4(O) + 36cos^2(O) - 2

Lets begin again with the LHS, note that:
2cos(6O) = 2cos(2(3O))......................(5)

We also know the identity:
cos(2O) = 2cos^2(O) - 1......................(6)

Now notice that (5) and (6) are essentially the same thing, we simply substitute 3O and not O into (6), and we will get (5). Do you see that?
2cos(2(3O)) = 2[2cos^2(3O) - 1]................(7)

Now we can not use the same trick on (7), because it is not a multiple of 2. What other trig identities do we know that might be able to help us? How about:
cos(A+B) = cosA.cosB - sinA.sinB......................(8)

we can re-write:
cos^2(3O) = cos^2(O+2O)

So using (8) this becomes:
cos^2(3O) = cos^2(O+2O) = [cosO.cos2O - sinO.sin2O]^2............(9)

In (9) we want to simplify cos2O and sin2O in terms of O (not 2O), because the RHS of question 2 (what we are trying to prove) is written in terms of O only, and not 2O. We know how to simplify cos2O from (6), and we know th identity:
sin2O = 2sinO.cosO....................(10)

So using (6) and (10) lets simplify (9):
cos^2(3O) = (cosO.cos2O - sinO.sin2O)^2
= (cosO.[2cos^2O - 1] - sinO.[2sinO.cosO])^2
= (2cos^3O - cosO - 2sin^2O.cosO)^2.............(11)

Now, the RHS of question 2 contains only cos's, so we want to get rid of the sin^2O in (11) and write it in terms of cos. So we use:
sin^2O + cos^2O = 1
Re-arrange:
sin^2O = 1 - cos^2O...........................(12)

We now use (12) to simplify (11):
cos^2(3O) = (2cos^3O - cosO - 2sin^2O.cosO)^2
= (2cos^3O - cosO - 2[1-cos^2O].cosO)^2
= (2cos^3O - cosO - 2cosO+ 2cos^3O)^2
= (4cos^3O-3cosO)^2
= 16cos^6O - 24cos^4O + 9cos^2O................(13)

Now substitute (13) into (7):
2[2cos^2(3O) - 1] = 4cos^2O - 2
= 4[16cos^6O - 24cos^4O + 9cos^2O] - 2
= 64cos^6O - 96cos^4O + 36cos^2O - 2

Thus if we follow the reasoning from (5) to (7) to the above, we find we have proven that LHS = RHS and so the identity is proven
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I want you to try the rest for yourself. The only way your mathsimproves is if you do problems yourself. You should always try to solve problems that are only just a bit too hard for you to do. It doesn't matter if you have to spend hours on one problem, because in the end, when you solve the problem, you will have a deeper understanding of the topic, and you grow mathematically.
The rest of the problem are solved similarly. Remember you always want to start by trying to symplify the shorter side, as done above. Sometimes you will need to simplify both sides, and show that the simplifications are equal.
Remeber to try and write everything in terms of the trig identities you already know. The rest is just algebra. Nothing fancy.

Hint for question 10:
(alpha) + (beta) + (gamma) = pi
take the tan of both sides, then try grouping the argument of tan.

If you need any more help, feel free to ask