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Question Number 4783 by Dnilka228 last updated on 10/Mar/16
cosα+β≈(cosα+cosβcos−1α+cos−1β)α+βsina+b≈(sina+sinbsin−1a+sin−1b)a+btan(a+ab)k≈(tan(a+b)×ktan−1(a+b)×k)
Commented by Dnilka228 last updated on 10/Mar/16
sin(a+ab)Γ≈(sin(a+b)×Γsin−1(a−b)÷Γ)cos(Γ=e+π2)α+β≈(cose+cosπ2cos−1e−cos−1π2)
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