Viète, Euler and Brahmagupta: Beauty of Trigonometry
NOTE: The level of Math here is easier than the previous article. For challenging stuff look through Viète and the Challenge Problems.
A lot of Math-enthusiasts dislike doing too much of geometry and perhaps that is justified because before high school,geometry is mostly drawing and measurements. However, what makes trigonometry different is the amount of algebra involved, encountered by students mostly in high school- right from mundane sine-cosine identities to infinite series and calculus. It is this fusion of algebra and geometry that makes trigonometry so interesting. This article discusses few contributions of great Mathematicians, namely François Viète, Leonhard Euler and Brahmagupta- all of whom proved some really “cool” formulas and identities in Trigonometry.
François Viète
François Viète was a French Mathematician who gave the world some really beautiful proofs. He is mentioned extensively in Trigonometric Delights by Eli Maor (a book I just finished reading recently). One of his biggest achievements was to provide a trigonometric solution to special cubic equations. I had given a challenge problem in my last article related to the same, and let me describe the method shortly.
Consider the depressed cubic equation x³+Hx+G=0
Let x = u*cos θ (u is some constant we don’t know yet)
=> u³cos³ θ+Hucos θ+G=0
Dividing both sides of the equation by u³/4 gives:
4cos³ θ + (4H/u²)cos θ + (4G/u³) = 0 (we assume that u is not zero since G is some non-zero constant in the case)
Using this we can find u if we attempt to coincide this with the identity -
cos(3θ)=4cos³θ- 3cosθ, we equate 4H/u² and minus 3 to get:
u = ±2 √-H/3 => cos (3θ) =(3G√3)/2H √H
Thus we can solve for θ and hence find x.(However here the method only works if -1 <(3G√3)/2H √H < 1)
Viete was so adept at such transformations of algebraic equations to trigonometric equations that once when the Dutch Ambassador to France challenged King Henry IV with a 45th degree equation, Viete came up with 23 solutions to the problem. This served him well especially because the ambassador boasted that not a single Frenchman could solve it. Let me describe the method below-
The equation was:
x⁴⁵-45x⁴³+945x⁴¹-12300x³⁹+……..+95634x⁵-3795x³+45x=c for some constant ‘c’.
Now let c=2sin(45θ), z=2sin(15θ), y=2sin(5θ), x=2sinθ
Using the sin3θ= 3sinθ-4sin³θ identity, we get c=6z-8z³(let it be equation 1) and then using the same identity z=6y-8y³ (let it be equation 2).
Also since sin5θ = 16sin⁵θ-20sin³θ+5sinθ, we get y=x⁵-5x³+5x (Equation 3)
Substituting equation 3 for y in equation 2 to get z in terms of x and then substituting this new equation for z in equation 1 to get c in terms of x gives us exactly the original equation! (Unfortunately, back then trigonometric ratios were taken to be positive so Viete could only find 23 solutions). The complete set of solutions is given by:
Another magnificent achievement of Viete was the discovery of the remarkable formula:
But how did Viete come up with this?
Viete discovered this in 1593 and in proving it he used a geometric argument based on the ratio of areas of regular polygons of n and 2n sides inscribed in the same circle. This marks a milestone in mathematical history, as noted by Eli Maor in his book, because it was the first time “ an infinite process was explicitly written as a succession of algebraic operations”.
Leonhard Euler
Not much can be said of Euler, who undoubtedly is one of the greatest of geniuses to have stepped on planet earth. The only person comparable to Euler is Carl Gauss, but Euler’s achievements are nothing short of transcendental. Notable achievements include the discovery of the number e and also the formula e^(π*i) = -1. Some of his less known work includes a unique trigonometric proof of Heron’s formula for Area of triangle.
This proof is in fact unique to Euler- the very idea of taking tangents of all half-angles in a triangle and taking radius of in-circle to derive one of the most popular mathematical formulas is just ingenious. Euler still has a lot to be credited for- this article is just too small a space to even include another of his humongous feats.
Brahmagupta
Brahmagupta was the first person to compute rules for dealing with zero and also one of the first people to provide a general solution (although incomplete) to quadratic equations. His work on Diophantine equations are well noted by mathematicians the world over. One of his remarkable achievements was the discovery of the formula for the area of a cyclic quadrilateral. What really makes it remarkable is its similarity to the Heron’s formula (reason being that Heron’s formula is actually a special case of Brahmagupta’s formula), and perhaps trigonometry provides with the easiest, if not the most elegant method to prove it. The formula goes as follows:
Where S is the semi perimeter and a,b,c,d the sides. Now let’s try to prove this using simple trigonometric identities. What makes a cyclic quadrilateral different from other shapes? Well, the obvious answer would be that the opposite angles add up to 180 degrees, and in fact, that is the most fundamental idea to proving this. Let’s build on this-
Notice how very simple identities of trigonometry like cosine rule and sine rule (for ares of triangles) lead to such an interesting result. As you will see in the challenge problem, this can be generalized for all quadrilaterals too.
Challenge Problems
Conclusion
Trigonometry has developed as field of Mathematics over the years. Initially it was limited to spherical shapes during Ancient Greece but later the world saw the development of flat-plane trigonometry. With this mathematicians began to find links between trigonometry and algebraic equations. Today, it has deep applications in engineering and astronomy. You’ll even encounter sines and cosines in complex analysis, differential geometry, etc.
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