/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 Show that if \(\theta \approx \f... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 59

Show that if \(\theta \approx \frac{\pi}{2}\), then \(\sin \theta \approx 1-\frac{1}{2}\left(\frac{\pi}{2}-\theta\right)^{2}\).

Short Answer

Expert verified
To show that when \(\theta \approx \frac{\pi}{2}\), the sine function can be approximated by \(1-\frac{1}{2}\left(\frac{\pi}{2}-\theta\right)^{2}\), we used the Taylor series expansion of \(\sin(\theta)\) around \(\frac{\pi}{2}\). After finding the derivatives and evaluating them at \(\frac{\pi}{2}\), we applied the Taylor series expansion formula and simplified the expression, resulting in \(\sin(\theta) \approx 1 - \frac{1}{2}\left(\frac{\pi}{2} - \theta\right)^2\).

Step by step solution

01

Taylor series expansion of sin(胃) around 蟺/2 #

To approximate the sine function around the angle of 蟺/2, we need to first find the Taylor series expansion of sin(胃) around 蟺/2. In general, the Taylor series expansion of a function f(x) around a point a is given by: \(f(x) = f(a) + f'(a)(x-a) + \frac{1}{2}f''(a)(x-a)^2 + \dots\) Now, let's find the derivatives of the sine function and evaluate them at 蟺/2: \(f(\theta) = \sin(\theta)\) \(f'(\theta) = \cos(\theta)\) \(f''(\theta) = -\sin(\theta)\) \(f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1\) \(f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0\) \(f''(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1\)
02

Applying the Taylor series expansion for sin(胃) #

Now that we have the derivatives of the sine function evaluated at 蟺/2, we can apply the Taylor series expansion formula to find the approximation for sin(胃) around 蟺/2: \(\sin(\theta) \approx 1 + 0 \cdot \left(\theta - \frac{\pi}{2}\right) - \frac{1}{2} \cdot (-1)\left(\theta - \frac{\pi}{2}\right)^2\) Simplifying the expression, we get: \(\sin(\theta) \approx 1 - \frac{1}{2}\left(\theta - \frac{\pi}{2}\right)^2\)
03

Conclusion #

We have successfully shown that when 胃 is close to 蟺/2, the sine function can be approximated by the given equation: \(\sin(\theta) \approx 1 - \frac{1}{2}\left(\frac{\pi}{2} - \theta\right)^2\)

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Most popular questions from this chapter

Show that $$(\cos \theta+\sin \theta)^{2}(\cos \theta-\sin \theta)^{2}+\sin ^{2}(2 \theta)=1$$ for all angles \(\theta\).

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