# 11.03 - Section 3.5: Inverse trig functions (concluded)....

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Section 3.5: Inverse trig functions (concluded). Last time we defined arcsin and arccos (also written as sin –1 and cos –1 ), and showed that ( d / dx ) arcsin x = 1 / sqrt(1– x 2 ) and ( d / dx ) arccos x = –1 / sqrt(1– x 2 ) for x in the interval … ..?. . (–1,1). If f ( x ) = cos –1 x + sin –1 x , what is f ( x )? . .?. . ..?. . (1 / sqrt(1– x 2 )) + (–1 / sqrt(1– x 2 )) = 0. What does this mean about f ( x )? . .?. . ..?. . It’s constant. What constant? ..?. . ..?. . Plug in x =0: f (0) = π /2 = f ( x ) for all x . General strategy : To show that f ( x ) = g ( x ) for all x , first show that f ( x ) = g ( x ) for all x ; this shows that f ( x ) and g ( x ) differ by a constant (once we know that the only functions with derivative equal to zero everywhere are

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constant functions; we’ll prove this in the next chapter). Then to nail down the constant, plug in a specific value of x (any x will do). In more detail: If f = g , then ( f g ) =0, so f g is constant, so there exists some C such that f ( x )= g ( x )+ C for all x ; if we can check that f ( x 0 ) = g ( x 0 ) for one particular x 0 , then we can conclude that C =0, so that f ( x )= g ( x ) for all x .
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## This note was uploaded on 02/13/2012 for the course MATH 141 taught by Professor Staff during the Fall '11 term at UMass Lowell.

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11.03 - Section 3.5: Inverse trig functions (concluded)....

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