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Use both the interpolation scheme of Brahmagupta and the algebraic formula of Bh?skara I to approximate \(\sin \left(16^{\circ}\right)\). Compare the two values to each other and to the exact value. What are the respective errors?

Short Answer

Expert verified
Answer: Brahmagupta's interpolation scheme provided a more accurate result for calculating the sine of 16 degrees, with an approximation error of approximately 0.001076, which is significantly smaller than the error of approximately 0.364000 for Bhāskara I's algebraic formula.

Step by step solution

01

Convert angle to radians

First, we need to convert the given angle from degrees to radians so that we can apply the interpolation scheme and algebraic formula more easily. To do this, use the conversion factor \(\frac{\pi}{180}\): $$16^{\circ} = 16 \cdot \frac{\pi}{180}$$ $$16^{\circ} = \frac{8\pi}{90}$$
02

Apply Brahmagupta's interpolation scheme

Brahmagupta's interpolation scheme provides an approximation of the sine of an angle by comparing it to the sines of the closest known integer multiples of \(15^{\circ}\). Since \(16^{\circ}\) is between \(15^{\circ}\) and \(30^{\circ}\), we can use this method: $$\sin (16^{\circ}) \approx \sin (15^{\circ}) + \frac{\sin (30^{\circ}) - \sin (15^{\circ})}{15}$$ Using some known sine values, we get: $$\sin (16^{\circ}) \approx \sin (15^{\circ}) + \frac{\frac{1}{2} - \sin (15^{\circ})}{15}$$ Now, calculate the approximate value: $$\sin (16^{\circ}) \approx 0.258819 + 0.016105$$ $$\sin (16^{\circ}) \approx 0.274924$$
03

Apply Bhāskara I's algebraic formula

Bhāskara I's algebraic formula is as follows: $$\sin \left(\theta\right) \approx \frac{16\theta \left(180 - \theta\right)}{5\theta^2 + 4 \left(180 - \theta\right)}$$ Plug \(\theta = 16^{\circ}\) into the formula: $$\sin \left(16^{\circ}\right) \approx \frac{16 \cdot 16 (180 - 16)}{5 \cdot 16^2 + 4 \cdot (180 - 16)}$$ Now, calculate the approximate value: $$\sin \left(16^{\circ}\right) \approx \frac{16 \cdot 16 (164)}{5 \cdot 16^2 + 4 \cdot 164}$$ $$\sin \left(16^{\circ}\right) \approx \frac{41984}{65600}$$ $$\sin \left(16^{\circ}\right) \approx 0.640000$$
04

Compare the approximations to the exact value

Calculate the exact value of \(\sin \left(16^{\circ}\right)\) and compare it to the approximations: Exact value: \(\sin \left(16^{\circ}\right) \approx 0.276^{\cdot \cdot \cdot}\) Approximation using Brahmagupta's interpolation scheme: \(0.274924\) Approximation using Bhāskara I's algebraic formula: \(0.640000\)
05

Determine the errors

Determine the errors of the approximations by subtracting them from the exact value: Error for Brahmagupta's interpolation scheme: \(|0.276^{\cdot \cdot \cdot} - 0.274924| \approx 0.001076\) Error for Bhāskara I's algebraic formula: \(|0.276^{\cdot \cdot \cdot} - 0.640000| \approx 0.364000\) The error for the approximation using Brahmagupta's interpolation scheme is significantly smaller than the error for the approximation using Bhāskara I's algebraic formula.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Brahmagupta's Interpolation Scheme
Brahmagupta's interpolation scheme is a method used in ancient mathematics to approximate trigonometric functions. This technique is particularly useful when you want an estimated value of the sine function for angles that fall between known values. For example, to find \( \sin(16^{\circ}) \), Brahmagupta's scheme takes advantage of the nearby known sine values at \( 15^{\circ} \) and \( 30^{\circ} \).
The formula used is:
  • \( \sin(\theta) \approx \sin(a) + \frac{\sin(b) - \sin(a)}{b-a} \times (\theta-a) \)
Here, \( a \) and \( b \) are angles with known sine values, and \( \theta \) is the angle for which you are trying to approximate the sine value. In the given example:
  • \( a = 15^{\circ} \)
  • \( b = 30^{\circ} \)
  • \( \sin(16^{\circ}) \approx \sin(15^{\circ}) + \frac{\sin(30^{\circ}) - \sin(15^{\circ})}{15} \)
Once the values are substituted and simplified, the approximation is obtained. It's a straightforward method focused on linear interpolation and provides reasonable accuracy for small angles.
Bhāskara I's Algebraic Formula
Bhāskara I, a notable mathematician from ancient India, introduced an algebraic formula to approximate the sine function. This formula is known for its simplicity and ingenuity. The expression attempts to create a polynomial approximation to predict sine values.
The formula is given by:
  • \( \sin(\theta) \approx \frac{16\theta(180-\theta)}{5\theta^2 + 4(180-\theta)} \)
In this formula:
  • \( \theta \) is in degrees.
To use Bhāskara's formula, you simply substitute the angle into this expression. For \( 16^{\circ} \), substitution gives an approximation.
However, it is essential to remember that the formula can exhibit larger errors depending on the angle used because it is optimized for small angles near zero or specific ranges. The calculated value might vary widely from the actual sine value, showcasing its limitations.
This technique beautifully merges algebraic operations with trigonometric functions, reflecting the rich mathematical heritage from Bhāskara's era.
Sine Function
The sine function is one of the core elements of trigonometry, often used in various fields such as physics, engineering, and mathematics. It describes the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. Its formula in terms of the unit circle is:
  • \( \sin(\theta) = \frac{opposite}{hypotenuse} \)
Beyond basic geometry, sine is treated as a periodic function, repeating its values every \( 360^{\circ} \) or \( 2\pi \) radians.
  • The sine wave has key characteristics like amplitude, period, frequency, and phase.
These properties make it versatile in modeling periodic phenomena.
When calculating sine values, especially for angles not on standard charts, approximation techniques like Brahmagupta's scheme or Bhāskara I's formula become quite useful. Tools like these were developed long before modern calculators, demonstrating the human pursuit of mathematical understanding and precision.
Error Analysis
Error analysis is crucial when approximating values, as seen in calculating \( \sin(16^{\circ}) \) using Brahmagupta's scheme and Bhāskara's formula. This process involves comparing the approximate value with the exact value and determining the deviation.
The exact sine value for \( 16^{\circ} \) is approximately \( 0.276 \). By calculating errors:
  • Brahmagupta's approximation is \( 0.274924 \), with an error of \( |0.276 - 0.274924| \approx 0.001076 \).
  • Bhāskara I's formula gives \( 0.640000 \), resulting in a larger error of \( |0.276 - 0.640000| \approx 0.364000 \).
Such analysis helps in understanding which method offers a closer approximation and better accuracy.
This step is vital for evaluating approximation methods, especially when precision is necessary. It allows mathematicians and scientists to select the best-suited technique for their specific applications, ultimately saving time and resources while ensuring dependable results.

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Most popular questions from this chapter

Why would the Indians have thought it better to use an algebraic approximation to the sine function rather than calculate values using geometric methods and methods of interpolation?

Prove that Brahmagupta's procedure does give a solution to the simultaneous congruences. Begin by noting that the Euclidean algorithm allows one to express the greatest common divisor of two positive integers as a linear combination of these integers. Note further that a condition for the solution procedure to exist is that this greatest common divisor must divide the "additive." Brahmagupta does not mention this, but Bh?skara and others do.

Show that if \((u, v)\) is a solution to \(D x^{2}+2=y^{2}\), then \(\left(u_{1}, v_{1}\right)=\left(u v, v^{2}-1\right)\) is a solution to \(D x^{2}+1=y^{2} .\) Deduce a similar rule if \((u, v)\) is a solution to \(D x^{2}-2=y^{2}\).

Show that Bh?skara's algebraic formula for the Sine can be rewritten as an approximation formula for the modern sine in the form $$ \sin x \approx \frac{16 x(\pi-x)}{5 \pi^{2}-4 x(\pi-x)} $$ where \(x\) is given in radians. Graph this function on a graphing calculator from 0 to \(\pi\) and compare it with the graph of \(\sin x\) on that interval.

Brahmagupta asserts that if \(A B C D\) is a quadrilateral inscribed in a circle, with side lengths \(a, b, c, d\) (in cyclic order) (see Fig. 8.8), then the lengths of the diagonals \(A C\) and \(B D\) are given by $$ A C=\sqrt{\frac{(a c+b d)(a d+b c)}{a b+c d}} $$ and similarly $$ B D=\sqrt{\frac{(a c+b d)(a b+c d)}{a d+b c}} $$ Prove this result as follows: a. Let \(\angle A B C=\theta\). Then \(\angle A D C=\pi-\theta\). Let \(x=A C\). Use the law of cosines on each of triangles \(A B C\) and \(A D C\) to express \(x^{2}\) two different ways. Then, since \(\cos (\pi-\theta)=-\cos \theta\), use these two formulas for \(x^{2}\) to determine \(\cos \theta\) as a function of \(a, b, c\), and \(d\). b. Replace \(\cos \theta\) in your expression for \(x^{2}\) in terms of \(a\) and \(b\) by the value for the cosine determined in part \(\mathrm{a}\). c. Show that \(c d\left(a^{2}+b^{2}\right)+a b\left(c^{2}+d^{2}\right)=(a c+b d)(a d+b c)\). d. Simplify the expression for \(x^{2}\) found in part \(\mathrm{b}\) by using the algebraic identity found in part c. By then taking square roots, you should get the desired expression for \(x=A C\). (Of course, a similar argument will then give you the expression for \(y=B D\).)

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