# C2 2 y tan x sin x cos x gives y cos x cos x sin x

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Chapter 4 / Exercise 16
Elementary and Intermediate Algebra: Algebra Within Reach
Larson
Expert Verified
C.2 2. y = tan x = sin x cos x gives y = cos x cos x sin x( sin x) cos 2 x = cos 2 x + sin 2 x cos 2 x = 1 + tan 2 x . Since cos 2 x + sin 2 x = 1, we get y = 1 / cos 2 x as an alternative answer. 4. (a) a sin ax (b) a sin bt + abt cos bt (c) a cos (at + b) sin[sin (at + b) ] cos { cos[sin (at + b) ] } 6. f (x) = 3 ( sin x x 1 ) 2 ( cos x 1 ) . In the open interval I = ( 0 , 3 π/ 2 ) , f (x) = 0 only at cos x = 1, when x = π/ 2. (It is easy to see that sin x < x + 1 for all x 0 because the function defined by g(x) = x + 1 sin x for all x 0 satisfies g( 0 ) = 1 and g (x) = 1 cos x 0 for all x 0.) Thus, the only possible extreme points are 0, π/ 2, and 3 π/ 2. Comparing the function values at these points, we find that f (x) has its maximum 1 at x = 0, and its minimum ( 2 + 3 π/ 2 ) 3 at x = 3 π/ 2. (The extreme-value theorem ensures that extreme points do exist.) 8. Implicit differentiation yields 1 · cos y + x( sin y)y y sin x y cos x = 0, so y = cos y y cos x sin x + x sin y . At (π, π/ 2), y = 1 / 2, so the equation for the tangent is y = x/ 2. 10. (a) cos x + C (b) π/ 2 0 cos x dx = π/ 2 0 sin x = sin (π/ 2 ) 0 = 1 (c) Integrating by parts yields I = sin 2 x dx = sin x( cos x) cos x( cos x) dx = − sin x cos x + cos 2 x dx = − sin x cos x + ( 1 sin 2 x) dx . Hence, I = − sin x cos x + x I + C . Solving for I gives I = − 1 2 ( sin x cos x x) + C 1 . (d) π 0 x cos x dx = π 0 x sin x π 0 sin x dx = 0 + π 0 cos x = cos π cos 0 = − 2. 12. (a) π/ 4 (b) π/ 2 (c) π/ 6 (d) π/ 3. (You can read all these values off from Table C.1.) 14. Make use of [C.20] and the chain rule to derive y = 1 1 + e x e x 2 2 · e x + e x 2 . Then simplify. C.3 2. See Fig. C.3.2. © Knut Sydsæter and Peter Hammond 2010
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Chapter 4 / Exercise 16
Elementary and Intermediate Algebra: Algebra Within Reach
Larson
Expert Verified
69 4. (a) 2 3 ( cos π/ 3 + i sin π/ 3 ) (b) cos π + i sin π (c) 4 ( cos 4 π/ 3 + i sin 4 π/ 3) (d) 2 ( cos 7 π/ 4 + i sin 7 π/ 4 ) Imaginary axis 2 i i i 2 i 3 i 4 i Real axis 1 1 2 3 4 w = 1 + 3 i z + w = 3 + i z = 2 2 i Figure C.3.2 © Knut Sydsæter and Peter Hammond 2010
70 TEST I (Elementary Algebra) A certain familiarity with elementary algebra is an essential prerequisite for reading the textbook (and for understanding most modern economics texts). This test is designed for students and instructors to discover whether the students have the proper background. (In a number of countries, many beginning economics students’ background in elementary algebra appears to have become much weaker during the last few years. In fact, lecturers using this test (or similar ones) have been shocked by the results, and have had to readjust their courses.) At the head of each problem, immediately after the number, the relevant sections of the introductory chapters in the book are given in parentheses, followed in square brackets by the number of points for a correct answer to each separate part of the problem. In a 20–30 minute test, any student who scores less than 50 (out of 100) has serious problems with elementary algebra. Such students definitely need to review the relevant section of Chapters 1 and 2, or consult other elementary material.