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Chapter 8: Problem 31

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.

Short Answer

Expert verified
Answer: Yes, Bhaskara's algebraic formula for the sine, rewritten as the given approximation formula, closely resembles the modern sine function in the interval from 0 to π.

Step by step solution

01

Rewrite Bhaskara's formula for sine as the given approximation formula

We are given Bhaskara's formula for the sine as follows: $$ \sin x \approx \frac{16 x(\pi-x)}{5 \pi^{2}-4 x(\pi-x)} $$ Our goal is to show that this formula indeed is an approximation for the modern sine function, i.e., \(\sin x\).
02

Select the interval

We are asked to consider the interval \(0\) to \(\pi\), where \(x\) is given in radians.
03

Graph both functions

Now, we need to graph both the functions \(\sin x\) and \(f(x) = \frac{16 x(\pi-x)}{5 \pi^{2}-4 x(\pi-x)}\) on the same graphing calculator in the given interval. In the graphing calculator, enter the two functions: 1. \(\sin x\) 2. \(\frac{16 x(\pi-x)}{5 \pi^{2}-4 x(\pi-x)}\) and graph them in the interval \([0, \pi]\).
04

Compare the graphs

After graphing both the functions, you will notice that the two graphs closely resemble each other. The graph of \(f(x) = \frac{16 x(\pi-x)}{5 \pi^{2}-4 x(\pi-x)}\) closely follows the graph of \(\sin x\) in the given interval. This shows that Bhaskara's algebraic formula for the sine can indeed be rewritten as an approximation formula for the modern sine in the form given and that the approximation is pretty accurate in the interval \([0, \pi]\).

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Key Concepts

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

Bhaskara's approximation
Bhaskara's approximation is a fascinating way to calculate the sine of an angle using an algebraic method rather than a more complex trigonometric approach. This formula is especially useful for approximations without the need for advanced mathematics.

The approximation can be expressed as:
  • \( \sin x \approx \frac{16 x(\pi-x)}{5 \pi^{2}-4 x(\pi-x)} \)
In this formula, \(x\) represents the angle in radians. This approach provides a reasonably accurate representation of the sine function for values of \(x\) within the range of \([0, \pi]\).

The mathematical roots of Bhaskara’s work trace back to his influential impact on the development of trigonometry in India. This particular approximation is notable for its ability to yield results that closely align with the sine function across a significant range without the need for a calculator.
Sine function
The sine function, one of the fundamental trigonometric functions, describes the relationship between an angle and the length of the opposite side of a right triangle. In modern terms, the sine function is defined as:
  • \( \sin x = \frac{\text{opposite side}}{\text{hypotenuse}} \)
In the unit circle, \( \sin x\) signifies the y-coordinate of a point on the circle, given an angle \(x\) from the positive x-axis.

Sine values range from -1 to 1, describing a wave-like pattern that repeats every \(2\pi\). This behavior is periodic, meaning it repeats at regular intervals, specifically at every \(2\pi\) radians or 360 degrees.

Applications of the sine function extend beyond basic trigonometry. It's crucial in physics for describing oscillations and waves, including sound and light waves. Understanding the basic wave properties enhanced by sine equations allows further exploration into wave mechanics and harmonic motions.
Graphing calculators
Graphing calculators are essential tools for visualizing mathematical concepts, particularly when examining trigonometric functions like sine. Utilizing a graphing calculator allows students and professionals alike to directly compare theoretical models with visual representations.

When exploring Bhaskara's approximation and the sine function within the interval \([0, \pi]\), input the formulas into a graphing calculator:
  • \( \sin x \)
  • \( \frac{16 x(\pi-x)}{5 \pi^{2}-4 x(\pi-x)} \)
This will yield two graphs that can be analyzed to understand the accuracy of Bhaskara's approximation.

Graphing these equations on the same axes reveals that the approximation closely matches the actual sine function, showcasing the practical use of this method in educational and real-world scenarios. Beyond educational settings, graphing calculators are also utilized in professional fields like engineering and architecture, where they play a pivotal role in designing and analyzing structures based on trigonometric principles. This snug fit on the graph emphasizes its effectiveness and the cleverness of Bhaskara's technique.

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Most popular questions from this chapter

Solve the following problem from Mah?vira: There are 4 pipes leading into a well. Among these, each fills the well (in order) in \(1 / 2,1 / 3,1 / 4\), and \(1 / 5\) of a day. In how much of a day will all of them together fill the well and each of them to what extent?

Brahmagupta asserts that if \(A B C D\) is a quadrilateral inscribed in a circle, as in Exercise 7 , then if \(s=\frac{1}{2}(a+\) \(b+c+d\) ), the area of the quadrilateral is given by \(S=\) \(\sqrt{(s-a)(s-b)(s-c)(s-d)}\) (Fig. 8.12). Prove this result as follows: Area of a quadrilateral inscribed in a circle a. In triangle \(A B C\), drop a perpendicular from \(B\) to point \(E\) on \(A C\). Use the law of cosines applied to that triangle to show that \(b^{2}-a^{2}=x(x-2 A E)\). b. Let \(M\) be the midpoint of \(A C\), so \(x=2 A M\). Use the result of part a to show that \(E M=\left(b^{2}-a^{2}\right) / 2 x\). c. In triangle \(A D C\), drop a perpendicular from \(D\) to point \(F\) on \(A C\). Use arguments similar to those in parts a and \(\mathrm{b}\) to show that \(F M=\left(d^{2}-c^{2}\right) / 2 x\). d. Denote the area of quadrilateral \(A B C D\) by \(P\). Show that \(P=\frac{1}{2} x(B E+D F)\) and therefore that \(P^{2}=\frac{1}{4} x^{2}(B E+\) \(D F)^{2} .\) e. Extend \(B E\) to \(K\) such that \(\angle B K D\) is a right angle, and complete the right triangle \(B K D\). Then \(B E+D F=\) \(B K\). Substitute this value in your expression from part d; then use the Pythagorean Theorem to conclude that \(P^{2}=\) \(\frac{1}{4} x^{2}\left(y^{2}-E F^{2}\right)\) f. Since \(E F=E M+F M\), conclude that \(E F=\left[\left(b^{2}+\right.\right.\) \(\left.\left.d^{2}\right)-\left(a^{2}+c^{2}\right)\right] / 2 x\). Substitute this value into the expression for \(P^{2}\) found in part e, along with the values for \(x^{2}\) and \(y^{2}\) found in Exercise 7. Conclude that $$ \begin{aligned} P^{2} &=\frac{1}{4}(a c+b d)^{2}-\frac{1}{16}\left[\left(b^{2}+d^{2}\right)-\left(a^{2}+c^{2}\right)\right]^{2} \\\ &=\frac{1}{16}\left(4(a c+b d)^{2}-\left[\left(b^{2}+d^{2}\right)-\left(a^{2}+c^{2}\right)\right]^{2}\right) \end{aligned} $$ g. Since \(s=\frac{1}{2}(a+b+c+d)\), show that \(s-a=\) \(\frac{1}{2}(b+c+d-a), s-b=\frac{1}{2}(a+c+d-b), s-c=\) \(\frac{1}{2}(a+b+d-c)\), and \(s-d=\frac{1}{2}(a+b+c-d)\) h. To prove the theorem, it is necessary to show that the final expression for \(P^{2}\) given in part \(\mathrm{f}\) is equal to the product of the four expressions in part \(\mathrm{g}\). It is clear that the denominators are both equal to 16 . To prove that the numerators are equal involves a lot of algebraic manipulation. Work carefully and show that the two numerators are in fact equal.

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 \(y=B D\).)

Another problem from Mah?v?ra: If 3 peacocks cost 2 coins, 4 pigeons cost 3 coins, 5 swans cost 4 coins, and 6 s?rasa birds cost 5 coins, and if you buy 72 birds for 56 coins, how many of each type of bird do you have?

Solve \(61 x^{2}+1=y^{2}\) by Bh?skara's process. The solution is \(x=226,153,980, y=1,766,319,049\).
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