Pre-Calc Homework Solutions 381

# Pre-Calc Homework Solutions 381 - Section 9.4 30(a sin2 x...

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30. (a) sin 2 x 5 } 1 2 } (1 2 cos 2 x ) 5 } 1 2 } 2 } 1 2 } 1 1 2 } (2 2 x ! ) 2 } 1 } (2 4 x ! ) 4 } 2 } (2 6 x ! ) 6 } 1 1 ( 2 1) n } ( ( 2 2 x n ) ) 2 ! n } 1 2 5 } 2 4 ? x 2 2! } 2 } 2 16 ? x 4 4 ! } 1 } 2 64 ? x 6 6 ! } 2 } 2 2 5 ? 6 x 8 8 ! } 1 } 1 2 0 ? 24 1 x 0 1 ! 0 } 2 5 x 2 2 } x 3 4 } 1 } 2 4 x 5 6 } 2 } 3 x 1 8 5 } 1 } 14 2 , x 1 1 7 0 5 } 2 (b) derivative 5 2 x 2 } 4 3 x 3 } 1 } 4 1 x 5 5 } 2 } 3 8 1 x 5 7 } 1 (c) part (b) 5 2 x 2 } (2 3 x ! ) 3 } 1 } (2 5 x ! ) 5 } 2 } (2 7 x ! ) 7 } 1 5 sin 2 x 31. (a) It works. For example, let n 5 2. Then P 5 3.14 and P 1 sin P < 3.141592653, which is accurate to more than 6 decimal places. (b) Let P 5 p 1 x where x is the error in the original estimate. Then P 1 sin P 5 ( p 1 x ) 1 sin ( p 1 x ) 5 p 1 x 2 sin x But by the Remainder Theorem, ) x 2 sin x ) , } ) x 6 ) 3 } . Therefore, the difference between the new estimate P 1 sin P and p is less than } ) x 6 ) 3 } . 32. (a)
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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