(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" (31/2) = -cos(3pi/2) Q(x), the quadratic approximation to cos(x) at a = 3/2 is: Q(x) =| Use Q(x) to approximate cos(4.6). cos(4.6)
(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" (31/2) = -cos(3pi/2) Q(x), the quadratic approximation to cos(x) at a = 3/2 is: Q(x) =| Use Q(x) to approximate cos(4.6). cos(4.6)
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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![**Approximate \(\cos(4.6)\) using a quadratic approximation:**
First, note that \(\cos(4.6) \approx \cos(3\pi/2)\).
Let \(f(x) = \cos(x)\). Then,
\[ f'(x) = -\sin(x) \]
and
\[ f''(x) = -\cos(x) \]
Let \(a = 3\pi/2\). Then
\[ f'(3\pi/2) = -\sin(3\pi/2) \]
and
\[ f''(3\pi/2) = -\cos(3\pi/2) \]
\(Q(x)\), the quadratic approximation to \(\cos(x)\) at \(a = 3\pi/2\) is:
\[ Q(x) = \ ]
Use \(Q(x)\) to approximate \(\cos(4.6)\).
\[
\cos(4.6) \approx
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Transcribed Image Text:**Approximate \(\cos(4.6)\) using a quadratic approximation:**
First, note that \(\cos(4.6) \approx \cos(3\pi/2)\).
Let \(f(x) = \cos(x)\). Then,
\[ f'(x) = -\sin(x) \]
and
\[ f''(x) = -\cos(x) \]
Let \(a = 3\pi/2\). Then
\[ f'(3\pi/2) = -\sin(3\pi/2) \]
and
\[ f''(3\pi/2) = -\cos(3\pi/2) \]
\(Q(x)\), the quadratic approximation to \(\cos(x)\) at \(a = 3\pi/2\) is:
\[ Q(x) = \ ]
Use \(Q(x)\) to approximate \(\cos(4.6)\).
\[
\cos(4.6) \approx
\]
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