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9₁¹ = = Σ m=3,5,7.... (2) sin() a 2 ( 2² h ²2 (1-m²)) 2ma² 2 4maa 0 =√√√(4) (3 + a ħ² (sin (3x) + 4maa =√(4m) = Pmº 0 1-25 95⁰ + 1-49 97⁰ + ...) ma =√√(a) (sin (2x) - h² Therefore, the first three terms are 4 sin (5x) + sin(x) + ....) sin(x)+sin(x) + ...) ma √(a) (sin (2x) - sin (²x) + ¹sin (25x) + ....). 2

9₁¹ = = Σ m=3,5,7.... (2) sin() a 2 ( 2² h ²2 (1-m²)) 2ma² 2 4maa 0 =√√√(4) (3 + a ħ² (sin (3x) + 4maa =√(4m) = Pmº 0 1-25 95⁰ + 1-49 97⁰ + ...) ma =√√(a) (sin (2x) - h² Therefore, the first three terms are 4 sin (5x) + sin(x) + ....) sin(x)+sin(x) + ...) ma √(a) (sin (2x) - sin (²x) + ¹sin (25x) + ....). 2

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.2: Trigonometric Functions Of Angles
Problem 81E
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Could you explain how we expanded the sum ? Why m got outside of the sum as a constant?

= a - = 4maa φι' = Σ π² ħ² m=3,5,7.... 4maa π² ħ² -) (sin (3x) + (2) sin a ma =√(²) (sin (²x) - π² ħ² 2ma² 0 √√² (1993° +1=2595° +1-4997° + ....) a (1 mл 2 - m² Pmº 4 sin (x) + sin(x) + sin(x) + ...) ma Therefore, the first three terms are √√(a) (sin (2x)-sin (2x) + ¹sin (25x) + ...). =) sin(x) + ....)
Transcribed Image Text:= a - = 4maa φι' = Σ π² ħ² m=3,5,7.... 4maa π² ħ² -) (sin (3x) + (2) sin a ma =√(²) (sin (²x) - π² ħ² 2ma² 0 √√² (1993° +1=2595° +1-4997° + ....) a (1 mл 2 - m² Pmº 4 sin (x) + sin(x) + sin(x) + ...) ma Therefore, the first three terms are √√(a) (sin (2x)-sin (2x) + ¹sin (25x) + ...). =) sin(x) + ....)
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