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Inverse trigonometric functions If we restrict the sine function to the interval [ − π2 , π2 ] , it becomes one-to-one; the inverse of this function is denoted by arcsin or sin −1 : (1, p2 ) domain: p 2 p 2 y = sin x – derivative: y = arcsin x range: [− 1, 1] [ − π2 , π2 ] ( 1, p2 ) d 1 arcsin x = dx 1 − x2 – properties: arcsin(sin x ) = x if x ∈ [ − π2 , π2 ] sin(arcsin x ) = x if x ∈ [ −1,1] If we restrict the cosine function to the interval [0, π] , it becomes one-to-one; the inverse of this function is denoted by arccos or cos −1 : y = cos x ( 1,p) y = arccos x p domain: range: [− 1, 1] [0, π] (1, 0) – derivative: d −1 arccos x = dx 1 − x2 – properties: arccos(cos x ) = x if x ∈ [ 0, π] cos(arccos x ) = x if x ∈ [ −1,1] If we restrict the tangent function to the interval ( − π2 , π2 ) , it becomes one-to-one; the inverse of this function is denoted by arctan or tan −1 : y = tan x p 2 domain: p 2 p 2 p 2 – derivative: 1 d arctan x = 1+ x2 dx – properties: arctan(tan x ) = x if x ∈ ( − π2 , π2 ) tan(arctan x ) = x y = arctan x range: (− ∞, ∞ ) ( − π2 , π2 ) Simplifying trig(trig −1 x) – First, draw a figure answering the question posed by the inverse function. Be sure to draw the triangle in the correct region, depending on the range of the inverse-trig function: – Determine the third side via the Pythagorean theorem (it will be positive). – Read off whatever trig function you need from the figure. For example: 3 5 q sin(arccos( −54 )) = 3 5 -4 −1 Simplifying trig (trig x ) : first evaluate the inside function, then answer the question posed by the inverse-trig function. Solving problems via trigonometric and inverse trigonometric functions: in general, you’re looking to find right triangles, from which you can read off trig relations. l’Hôpital’s rule f ( x) gives either 0 0 or ∞ ∞ , x →… g ( x ) f ′( x ) and if (2) lim exists or is infinite, x →… g ′( x ) then the original limit does the same thing. l’Hôpital’s rule states that if: (1) lim – Note that both hypotheses must be true for the rule to apply, and that the logic only works right-to-left; don’t just carelessly write “=”! 1 ∞ Other applications of l’Hôpital’s rule: 0 ⋅ ∞ ; ∞ − ∞ ; 0 , 1 , ∞ 0 – 0 ⋅ ∞ : first turn the product into a quotient by inverting one factor and moving it to the denominator. – ∞ − ∞ : first use algebra to turn this into a single quotient. 0 – 0 0 , 1∞ , ∞ :turn this into a natural exponential via a b = eb⋅ln a , take the limit of the exponent, and then use the limit of the exponent to find the original limit.