Sunday, 10 March 2013

algebra precalculus - How to I approach this problem?




Trigonometric or Algebraic computations? The radicals as denominator is not all that encouraging too and I don't see any promising method to evaluate it.





Solve for n in the equation below:
sin(902n)2+2+2+2n radicals=1+22



Answer



Use the half-angle formula:
cosx=2cos2x21
So
2+2cosx=4cos2x2
Then, for 0xπ2,

2cosx2=2+2cosx
So
2cosx2k+1=2+2cosx2k=2+2+2cosx2k1=



We have the denominator
2+2+2+2n radicals



and we know that cosπ=0, so
2cosπ2n+1=2+2+2+2n radicals




Hence, the LHS of the expression becomes
sin(902n)2+2+2+2n radicals=sin(π2n+1)2cos(π2n+1)=12tanπ2n+1



So we evaluate
12tanπ2n+1=1+22
which simplifies to
tanπ2n+1=21



n can be easily found from here.


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