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By Mejlbro L.

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Com 49 Calculus Analyse c1- 2 The Arcus Functions An alternative solution. e. 2 2 2 Arcsin sin π −x 2 = π − x, 2 for x ∈ [0, π]. From sin π π π − x = sin · cos(−x) + cos · sin(−x) = cos x, 2 2 2 we get Arcsin(cos x) = π − x, 2 for x ∈ [0, π]. 2) If x ∈ [−π, 0], then −x ∈ [0, π]. e. Arcsin(cos x) = π +x 2 for x ∈ [−π, 0]. 3) As a conclusion we have Arcsin(cos x) = π − |x| 2 for x ∈ [−π, π]. 5 Prove that cos(2 Arcsin x) = 1 − 2x2 , x ∈ [−1, 1]. A. Prove a formula. D. Apply the formula for cos 2y.

Similarly we see that x = cosh at and x = sinh at are both linearly independent solutions of the differential equation d2 x = a2 x, dt2 hence this equation has the complete solution x = c1 cosh at + c2 sinh at, t ∈ R, c1 , c2 ∈ R arbitrary constanter. Remark 1. We have implicitly above assumed that a > 0 (or just a = 0). For completeness we mention that if a = 0, then the complete solution is x = c1 + c2 t, which follows from two to successive integrations of d2 x = 0. com 52 Calculus Analyse c1- 2 The Arcus Functions Remark 2.

The corresponding heuristic calculation (which actually gives the right result, though the argument itself is mathematically wrong) is dx 1 1 1 = = . = −1 dy dy f (x) f (f (y)) dx ♦ 3) If y = F (x) = f (g(x)), then x = F −1 (y) = g −1 f −1 (y) . 4) First we note that according to (1), F (x) = f (g(x)) · g (x). An insertion into the result of (2) gives F −1 (y) = 1 1 = . 1 Find the derivative of y = Arccos x. A. The derivative of a given inverse function. D. g. the formula in a table. Then prove this formula.

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