mreditor16
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Sorry that was a 'Find' question, not 'Show that', editedShow that equals what sorry?
Squaring both and adding gives you 2*(expanded form of cos(a-b)) + 2 =8/3 (since sin^2a+cos^2a=1)
Squaring the two equations:
let that cube root thingo be a.
Gonna skip lots of lines, especially in the earlier parts, because this is extremely long.
From Newington Trials 2009
Got a difficult one for you guys. I saw that it was going to be in carrot's trial.
naa the answer is C.
equivalent of letting x = 2n-1, where n is any integer.
alternatively u can take the approach that every odd number is congruent toequivalent of letting x = 2n-1, where n is any integer.
let x = 2n-1 where n is an integer.
same shit to prove it for all negative odd but do n = k-1
but x^2 - 1 is an even function so it's sorta redundant to do it for negative numbers too
but it's still not proven for all odd, just shown for the first 4alternatively u can take the approach that every odd number is congruent to
1 mod 8
3 mod 8
5 mod 8
7 mod 8
Squaring it becomes
1 mod 8
9 mod 8 or 1 mod 8
25 mod 8 or 1 mod 8
49 mod 8 or 1 mod 8
Subtracting 1 from each gives 0 mod 8 i.e. all are divisible by 8
How so? Doesn't every odd number fit into those congruencies listed?but it's still not proven for all odd, just shown for the first 4
oh true yeah.How so? Doesn't every odd number fit into those congruencies listed?
haha it's okoh true yeah.
derp moment T_T