3 sin x 4 sin 3 x 2 sin x 2 sin 3 x sin x 2 sin 3 x 2

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Elementary Algebra
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Chapter 1 / Exercise 131
Elementary Algebra
Tussy/Gustafson
Expert Verified
3 sin x 4 sin 3 x 2 sin x 2 sin 3 x sin x 2 sin 3 x 2 sin x 1 sin 2 x sin x 2 sin 3 x 2 sin x cos 2 x sin x 2 sin 3 x 2 sin x cos x cos x 1 2 sin 2 x sin x sin 2 x cos x cos 2 x sin x sin 3 x sin 2 x x Power-Reducing Formulas The double-angle formulas can be used to obtain the following power-reducing formulas. Power-Reducing Formulas (See the proofs on page 405.) tan 2 u 1 cos 2 u 1 cos 2 u cos 2 u 1 cos 2 u 2 sin 2 u 1 cos 2 u 2 Example 5 Reducing a Power Rewrite as a sum of first powers of the cosines of multiple angles. Solution Property of exponents Power-reducing formula Expand binomial. Power-reducing formula Distributive Property Simplify. Factor. Now try Exercise 23. 1 8 3 4 cos 2 x cos 4 x 3 8 1 2 cos 2 x 1 8 cos 4 x 1 4 1 2 cos 2 x 1 8 1 8 cos 4 x 1 4 1 2 cos 2 x 1 cos 4 x 2 1 4 1 2 cos 2 x cos 2 2 x 1 cos 2 x 2 2 sin 4 x sin 2 x 2 sin 4 x STUDY TIP Power-reducing formulas are often used in calculus. Example 5 shows a typical power reduction that is used in calculus. Note the repeated use of power-reducing formulas.
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Elementary Algebra
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Chapter 1 / Exercise 131
Elementary Algebra
Tussy/Gustafson
Expert Verified
Half-Angle Formulas You can derive some useful alternative forms of the power-reducing formulas by replacing with The results are called half-angle formulas. u 2. u 390 Chapter 5 Analytic Trigonometry Half-Angle Formulas The signs of and depend on the quadrant in which lies. u 2 cos u 2 sin u 2 tan u 2 1 cos u sin u sin u 1 cos u cos u 2 ± 1 cos u 2 sin u 2 ± 1 cos u 2 Use your calculator to verify the result obtained in Example 6. That is, evaluate and You will notice that both expressions yield the same result. 2 3 2. sin 105 T E C H N O L O G Y T I P STUDY TIP To find the exact value of a trigonometric function with an angle in form using a half-angle formula, first convert the angle measure to decimal degree form. Then multiply the angle measure by 2. D M S Example 6 Using a Half-Angle Formula Find the exact value of Figure 5.30 Solution Begin by noting that 105 is half of 210 . Then, using the half-angle formula for and the fact that 105 lies in Quadrant II (see Figure 5.30), you have The positive square root is chosen because is positive in Quadrant II. Now try Exercise 39. sin 1 3 2 2 2 3 2 . 1 cos 30 2 sin 105 1 cos 210 2 sin u 2 105 ° 210 ° x y sin 105 .
Section 5.5 Multiple-Angle and Product-to-Sum Formulas 391 Example 7 Solving a Trigonometric Equation Find all solutions of in the interval 0, 2 . 2 sin 2 x 2 cos 2 x 2 Algebraic Solution Write original equation. Half-angle formula Simplify. Simplify. Pythagorean identity Simplify. Factor. By setting the factors and equal to zero, you find that the solutions in the interval are and Now try Exercise 57. x 0. x 3 2 , x 2 , 0, 2 cos x 1 cos x cos x cos x 1 0 cos 2 x cos x 0 2 1 cos 2 x 1 cos x 2 sin 2 x 1 cos x 2 sin 2 x 2 1 cos x 2 2 sin 2 x 2 ± 1 cos x 2 2 2 sin 2 x 2 cos 2 x 2 Graphical Solution Use a graphing utility set in radian mode to graph as shown in Figure 5.31. Use the zero or root feature or the zoom and trace features to approximate the -intercepts in the interval to be and These values are the approximate solutions of in the interval Figure 5.31

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