now that cos(ak – a/2) 2 sin(a/2) E sin(ak) : (1 %3D Taking the difference of cos(ak), A cos(ak) = -2 sin(a/2) sin(ak + a/2), (1 d replacing k by k – 12 gives A cos(ak – a/2) = -2 sin(a/2) sin(ak). (1 ow dividing by –-2 sin(a/2) and applying the operator A-1 gives cos(ak – a/2) 2 sin(a/2) A-1 sin(ak) = (1 hich, up to a constant, is equation (1.269).

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 40E
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1.8.5 Example E
Show that
cos(ak – a/2)
2 sin(a/2)
E sin(ak)
(1.269)
Taking the difference of cos(ak),
A cos(ak) = -2 sin(a/2) sin(ak + a/2),
(1.270)
and replacing k by k – 1/2 gives
A cos(ak – a/2) = -2 sin(a/2) sin(ak).
(1.271)
Now dividing by -2 sin(a/2) and applying the operator A-1 gives
cos(ak – a/2)
2 sin(a/2)
A-1 sin(ak) :
(1.272)
which, up to a constant, is equation (1.269).
Transcribed Image Text:1.8.5 Example E Show that cos(ak – a/2) 2 sin(a/2) E sin(ak) (1.269) Taking the difference of cos(ak), A cos(ak) = -2 sin(a/2) sin(ak + a/2), (1.270) and replacing k by k – 1/2 gives A cos(ak – a/2) = -2 sin(a/2) sin(ak). (1.271) Now dividing by -2 sin(a/2) and applying the operator A-1 gives cos(ak – a/2) 2 sin(a/2) A-1 sin(ak) : (1.272) which, up to a constant, is equation (1.269).
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