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use a linear approximation to estimate the following quantity. choose a…

Question

use a linear approximation to estimate the following quantity. choose a value of a that produces a small error and does not require the use of a calculator
cos 28°

the linear approximation of cos 28° is
(type an exact answer)

Explanation:

Step1: Convert degrees to radians

We know that \(x = 28^{\circ}=\frac{28\pi}{180}=\frac{7\pi}{45}\) radians. A good choice for \(a\) is \(30^{\circ}=\frac{\pi}{6}\) radians since \(\frac{\pi}{6}\approx0.5236\) and \(\frac{7\pi}{45}\approx0.4887\), and we know the exact value of \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\) and \(\sin(\frac{\pi}{6})=\frac{1}{2}\). The formula for linear approximation is \(L(x)=f(a)+f^{\prime}(a)(x - a)\), where \(f(x)=\cos(x)\), \(f^{\prime}(x)=-\sin(x)\)

Step2: Calculate \(L(x)\)

Substitute \(a = \frac{\pi}{6}\) and \(x=\frac{7\pi}{45}\) into the linear - approximation formula.
\(x-a=\frac{7\pi}{45}-\frac{\pi}{6}=\frac{14\pi - 15\pi}{90}=-\frac{\pi}{90}\)
\(f(a)=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\), \(f^{\prime}(a)=-\sin(\frac{\pi}{6})=-\frac{1}{2}\)

\(L(x)=\cos(\frac{\pi}{6})-\sin(\frac{\pi}{6})(\frac{7\pi}{45}-\frac{\pi}{6})\)
\(L(x)=\frac{\sqrt{3}}{2}-\frac{1}{2}(-\frac{\pi}{90})\)
\(L(x)=\frac{\sqrt{3}}{2}+\frac{\pi}{180}\)

Answer:

\(\frac{\sqrt{3}}{2}+\frac{\pi}{180}\)