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n-i-n-i-i-sin-n-i-




Question Number 4785 by Dnilka228 last updated on 10/Mar/16
n^i (n/(i×i))≈sin n^i +α
$${n}^{{i}} \frac{{n}}{{i}×{i}}\approx\mathrm{sin}\:{n}^{{i}} +\alpha \\ $$
Commented by Dnilka228 last updated on 10/Mar/16
n^i (n/(i×i))≠sin α+n^i   (√n^i )=sin n+sin i×(√2)  sin n^i ×f(x)=a+b=1,5  a=1, b=0,5
$${n}^{{i}} \frac{{n}}{{i}×{i}}\neq\mathrm{sin}\:\alpha+{n}^{{i}} \\ $$$$\sqrt{{n}^{{i}} }=\mathrm{sin}\:{n}+\mathrm{sin}\:{i}×\sqrt{\mathrm{2}} \\ $$$$\mathrm{sin}\:{n}^{{i}} ×{f}\left({x}\right)={a}+{b}=\mathrm{1},\mathrm{5} \\ $$$${a}=\mathrm{1},\:{b}=\mathrm{0},\mathrm{5} \\ $$
Commented by Dnilka228 last updated on 10/Mar/16
If a=1, sin a=sin 1  If b=0,5, sin b=sin 0,5
$${If}\:{a}=\mathrm{1},\:{sin}\:{a}={sin}\:\mathrm{1} \\ $$$${If}\:{b}=\mathrm{0},\mathrm{5},\:{sin}\:{b}={sin}\:\mathrm{0},\mathrm{5} \\ $$

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