(1 point) Approximate cos(4.6) using a quadratic approximat First note that cos(4.6) cos(3/2). Let f(x) = cos(x). Then, f'(x) = -sinx and f"(x) = -cosx Let a = 3x/2. Then f' (3/2) = -sin(3pi/2) and f" (3л/2) = -cos(3pi/2) Q(x), the quadratic approximation to cos(x) at a = 3/2 is Q(x) = cos(3pi/2) Use Q(x) to approximate cos(4.6). cos(4.6) cos(3pi/2)
(1 point) Approximate cos(4.6) using a quadratic approximat First note that cos(4.6) cos(3/2). Let f(x) = cos(x). Then, f'(x) = -sinx and f"(x) = -cosx Let a = 3x/2. Then f' (3/2) = -sin(3pi/2) and f" (3л/2) = -cos(3pi/2) Q(x), the quadratic approximation to cos(x) at a = 3/2 is Q(x) = cos(3pi/2) Use Q(x) to approximate cos(4.6). cos(4.6) cos(3pi/2)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![(1 point) Approximate cos(4.6) using a quadratic approximation:
First note that cos(4.6) cos(3/2).
Let f(x) = cos(x). Then,
f'(x) = -sinx
and
f"(x) = -cosx
Let a = 3x/2. Then
f' (3/2) = -sin(3pi/2)
and
f" (3л/2) = -cos(3pi/2)
Q(x), the quadratic approximation to cos(x) at a = 3/2 is:
Q(x) = cos(3pi/2)
Use Q(x) to approximate cos(4.6).
cos(4.6)
cos(3pi/2)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58cffe43-d701-4c21-9740-08fe8d98ee79%2Fabd1291f-b677-4379-b354-0d0be7e1faf9%2Fcdw2u49_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(1 point) Approximate cos(4.6) using a quadratic approximation:
First note that cos(4.6) cos(3/2).
Let f(x) = cos(x). Then,
f'(x) = -sinx
and
f"(x) = -cosx
Let a = 3x/2. Then
f' (3/2) = -sin(3pi/2)
and
f" (3л/2) = -cos(3pi/2)
Q(x), the quadratic approximation to cos(x) at a = 3/2 is:
Q(x) = cos(3pi/2)
Use Q(x) to approximate cos(4.6).
cos(4.6)
cos(3pi/2)
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