1873_solutions

X n 0 whenever n is a positive integer then the

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Unformatted text preview: 2 Solution: We have already seen that the restriction of the function sin to the interval /2, /2 is strictly increasing and that the range of this function is the interval 1, 1 . The function arcsin is therefore a strictly increasing continuous function from 1, 1 onto /2, /2 and, from the facts about differentiation of inverse functions studied earlier we know that for every number u between 1 and 1 we have 1 arcsin u  cos arcsin u Since the function cos is nonnegative on the interval /2, /2 we deduce that 1 1 1   arcsin u  cos arcsin u 1 u2 1 sin 2 arcsin u Therefore whenever 1  u  1 we have arcsin u  Þ0 u 1 1 t2 dt. 4. By analogy with the preceding exercise, give a definition of the function arctan and deduce that if u is any real number then u 1 dt arctan u  Þ 0 1  t2 The function tan is, of course, defined by the equation sin tan x  cos x x whenever cos x 0. At any such number x we have 1 0 tan x  cos 2 x  , is a strictly and we conclude that the restriction of the function tan to the interval 22 increasing function whose derivative is everywhere positive. Furthermore, since sin lim tan x  lim cos x  Ý x x /2 x /2 and x sin lim tan x  lim cos x  Ý, x /2 x /2  the range of the restriction of tan to , is the entire set R. 22  , . We deduce from the We define arctan to be the inverse function of the restriction of tan to 22 theorem on differentiation of inverse functions that if x is any real number we have 1 1 1 1. arctan x     1 tan arctan x 1  tan 2 arctan x 1  x2 2 cos arctan x It follows from the fundamental theorem that if u is any real number we have 379 arctan u  Þ0 u 1 dt. 1  t2 5. Prove that the restriction of the function cos to the interval 0,  is a strictly decreasing function from the interval 0,  onto the interval 1, 1 . Prove that if the function arccos is now defined to be the inverse function of this restriction of cos then for every number u 1, 1 we have  Þu 1 arccos u  dt. 2 0 1 t2 We know that cos 0  1 and cos   1 and that cos x  sin x  0 whenever 0  x  . Therefore cos is a strictly decreasing differentiable function from the interval 0,  onto 1, 1 whose derivative is positive at every number between 1 and 1. We define arccos to be the inverse function of the restriction of the function cos to 0,  . Note that because sin u 0 whenever 0 u  we have the identity sin arccos x  whenever 1 x 1  x  1 then cos 2 arccos x 1  1 x2 1. We deduce from the theorem on differentiation of inverse functions that if arccos x  1 cos arccos x 1 sin arccos x   1 1 x2 and it follows from the fundamental theorem that if 1  u  1 then u 1 arccos u  arccos 0  Þ dt 0 1 t2 u 1  Þ dt. 2 0 1 t2 6. Prove that if x and y are real numbers that are not both zero and if   arccos x2 x  y2 then sin   y . x2  y2 Deduce that if x and y are real numbers that are not both zero then there exists a positive number r and a real number  0, 2 such that x  r cos  and y  r sin . Suppos...
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