2 Use the identity sin x= 1- cos (2x) 2 to obtain the Maclaurin series for sin ´x. Then differentiate this series to obtain the Maclaurin series for 2 sin x cos x. Check that this is the series for sin (2x). What is the Maclaurin series for sin ²x? 00 Ο Α. Σ k=0 00 Ο Β. Σ k=0 00 ⭑C. Σ k=0 -- (-1)*22k+1 (2k+1)! (-1)*22k (2k)! x 2k (-1)*22k+1 00 (2k+2)! 1 * D. Σ 2 k=0 (-1)*22k (2k + 1)! 25 27 2k+1 3 5 7 9 =2x x + x 5! x + 22 2 =1- x + 4 x 2k+2 2 23 4 = 1 2k+1 2º x + 6 8 8 x 25 27 x + x 6 8! 8 -X +... ==-x+ -x 3! 3 - 6! 24 5 X + 2° 7! What is the Maclaurin series for 2 sin x cos x = sin (2x)? 00 O A. Σ k=0 00 'B. Σ k = 0 00 ○ C. Σ k = 0 00 OD. Σ k=0 2k (-1)+12 (2k)! (-12-1 (2k+1)! (-12-1 (2k)! 2k x =-1+ x x 2k+1 22 24 2 4 x + 6 X 2! =2x 23 3 7' 3' 3! + 25 X 5 - 7! 2k 21 (-1)+122k+2 (2k+1)! 2k+1 + 25 * 6 + 7 7 28 9 28 음... 8! 8 + ... -- =- 3 --- 5! + 7 7!

2 Use the identity sin x= 1- cos (2x) 2 to obtain the Maclaurin series for sin ´x. Then differentiate this series to obtain the Maclaurin series for 2 sin x cos x. Check that this is the series for sin (2x). What is the Maclaurin series for sin ²x? 00 Ο Α. Σ k=0 00 Ο Β. Σ k=0 00 ⭑C. Σ k=0 -- (-1)22k+1 (2k+1)! (-1)22k (2k)! x 2k (-1)22k+1 00 (2k+2)! 1 D. Σ 2 k=0 (-1)22k (2k + 1)! 25 27 2k+1 3 5 7 9 =2x x + x 5! x + 22 2 =1- x + 4 x 2k+2 2 23 4 = 1 2k+1 2º x + 6 8 8 x 25 27 x + x 6 8! 8 -X +... ==-x+ -x 3! 3 - 6! 24 5 X + 2° 7! What is the Maclaurin series for 2 sin x cos x = sin (2x)? 00 O A. Σ k=0 00 'B. Σ k = 0 00 ○ C. Σ k = 0 00 OD. Σ k=0 2k (-1)+12 (2k)! (-12-1 (2k+1)! (-12-1 (2k)! 2k x =-1+ x x 2k+1 22 24 2 4 x + 6 X 2! =2x 23 3 7' 3' 3! + 25 X 5 - 7! 2k 21 (-1)+122k+2 (2k+1)! 2k+1 + 25 6 + 7 7 28 9 28 음... 8! 8 + ... -- =- 3 --- 5! + 7 7!

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.3: Trigonometric Functions Of Real Numbers

Problem 24E

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