TAILIEUCHUNG - Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 68

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 68. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | Solving Trigonometric Equations 651 8. Let x cos x and g x arctan x. Find the following where a and b are positive constants. Your answers should be exact and as simple as possible. a g b T In Problems 9 through 11 simplify the expressions given that x e 0 y . 9. a sin-1 sin x 10. a sin-1 sin -x 11. a tan-1 tan x b cos-1 cos x b cos-1 cos -x b tan-1 tan -x In Problems 12 through 14 simplify the expressions given that x e y 2 . 12. a arcsin sin x b arccos cos x 13. a arcsin sin -x b arccos cos -x 14. a arctan tan x b arctan tan -x SOLVING TRIGONOMETRIC EQUATIONS A trigonometric equation is an equation in which the variable to be solved for is the argument of a trigonometric function. If we can get the trigonometric equation into the form sin u fc cos u fc or tan u fc where k is a constant then we can use the inverse trigonometric functions to help solve for u. EXAMPLE SOLUTION Solve for x. 4 cos x 1 -1 4 cos x 1 -1 4 cos x -2 1 cos x 2 One solution to this equation is cos 1 Ç- J. But there are actually two solutions for x e 0 and infinitely many solutions due to the periodicity of the cosine function. For this problem we can turn to a triangle we know and love. We know cos 3 . To have a negative cosine the angle must be in the second or third quadrant. So the solutions are 2ä 4ä x -3 f 2äm or x -3 f 2äm where n is any integer. See Figure on the following page for illustration. 652 CHAPTER 20 Trigonometry Circles and Triangles EXAMPLE Figure Solve for 0. 2 sin2 0 - sin 0 1 SOLUTION This is a quadratic equation in sin 0. We can either work with the sin 0 or we can begin the problem with the substitution u sin 0 solve for u and then return to sin 0. We ll take this latter approach. 2m2 u 1 2m2 u 1 0 u 1 2u 1 0 u 1 0 or u 1 or 2u 1 0 sin 0 1 or 1 sin 0 2 1 u 2 So 0 - 2nn n 5n or 0 -----F 2nn or 0 --------1- 2nn 6 6 where n is an integer. Figure Solving Trigonometric Equations 653 EXAMPLE SOLUTION EXAMPLE .

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