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(90∘2n)√2+√2+√2+…√2⏟n radicals=−1+√22
Answer
Use the half-angle formula:
cosx=2cos2x2−1
So
2+2cosx=4cos2x2
Then, for 0≤x≤π2,
2cosx2=√2+2cosx
So
2cosx2k+1=√2+2cosx2k=√2+√2+2cosx2k−1=…
We have the denominator
√2+√2+√2+…√2⏟n radicals
and we know that cosπ=0, so
2cosπ2n+1=√2+√2+√2+…√2⏟n radicals
Hence, the LHS of the expression becomes
sin(90∘2n)√2+√2+√2+…√2⏟n radicals=sin(π2n+1)2cos(π2n+1)=12tanπ2n+1
So we evaluate
12tanπ2n+1=−1+√22
which simplifies to
tanπ2n+1=√2−1
n can be easily found from here.
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