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Transcendental Functions
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birth and is locked in at 11% interest compounded continuously How much principal should be put into the account to yield the desired payoff
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SOLUTION Let P be the initial principal deposited in the account on the day of the nephew s birth Using our compound interest equation ( ), we have 100000 = P e11 21/100 , expressing the fact that after 21 years at 11% interest compounded continuously we want the value of the account to be \$100,000 Solving for P gives P = 100000 e 011 21 = 100000 e 231 = 992613 The aunt needs to endow the fund with an initial \$992613 You Try It: Suppose that we want a certain endowment to pay \$50,000 in cash ten years from now The endowment will be set up today with \$5,000 principal and locked in at a xed interest rate What interest rate (compounded continuously) is needed to guarantee the desired payoff
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Inverse Trigonometric Functions
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661 INTRODUCTORY REMARKS
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Figure 614 shows the graphs of each of the six trigonometric functions Notice that each graph has the property that some horizontal line intersects the graph at least twice Therefore none of these functions is invertible Another way of seeing this point is that each of the trigonometric functions is 2 -periodic (that is, the function repeats itself every 2 units: f (x + 2 ) = f (x)), hence each of these functions is not one-to-one If we want to discuss inverses for the trigonometric functions, then we must restrict their domains (this concept was introduced in Subsection 185) In this section we learn the standard methods for performing this restriction operation with the trigonometric functions
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INVERSE SINE AND COSINE
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Consider the sine function with domain restricted to the interval [ /2, /2] (Fig 615) We use the notation Sin x to denote this restricted function Observe that d Sin x = cos x > 0 dx
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CHAPTER 6 Transcendental Functions
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2 sec x 1 cos x _3 _2 _1 sin x
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_1 csc x _2
tan x
cot x
Fig 614
y = Sin x 1
Fig 615
Transcendental Functions
on the interval ( /2, /2) At the endpoints of the interval, and only there, the function Sin x takes the values 1 and +1 Therefore Sin x is increasing on its entire domain So it is one-to-one Furthermore the Sine function assumes every value in the interval [ 1, 1] Thus Sin : [ /2, /2] [ 1, 1] is one-to-one and onto; therefore f (x) = Sin x is an invertible function We can obtain the graph of Sin 1 x by the principle of re ection in the line y = x (Fig 616) The function Sin 1 : [ 1, 1] [ /2, /2] is increasing, one-to-one, and onto
y = Sin
_ F/2
Fig 616
The study of the inverse of cosine involves similar considerations, but we must select a different domain for our function We de ne Cos x to be the cosine function restricted to the interval [0, ] Then, as Fig 617 shows, g(x) = Cos x is a one-toone function It takes on all the values in the interval [ 1, 1] Thus Cos : [0, ] [ 1, 1] is one-to-one and onto; therefore it possesses an inverse We re ect the graph of Cos x in the line y = x to obtain the graph of the function Cos 1 The result is shown in Fig 618
EXAMPLE 635
Calculate
, Sin 3
, Cos
CHAPTER 6 Transcendental Functions
y = Cos x
Fig 617
y = Cos
Fig 618
SOLUTION We have Sin
3 2
, 3
Sin 1 0 = 0,
Transcendental Functions
= 4
2 2
Notice that even though the sine function takes the value 3/2 at many different values of the variable x, the function Sine takes this value only at x = /3 Similar comments apply to the other two examples We also have 5 3 1 Cos = , 2 6 Cos 1 0 = , 2 2 Cos 1 = 2 4 We calculate the derivative of f (t) = Sin 1 t by using the usual trick for inverse functions The result is d 1 1 (Sin 1 (x)) = = dx 1 x2 1 sin2 (Sin 1 x) The derivative of the function Cos 1 t is calculated much like that of Sin 1 t We nd that d 1 (Cos 1 (x)) = dx 1 x2 EXAMPLE 636