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Fearless Trigonometry – The Pythagorean Identities

The famous Pythagorean theorem extends to trigonometry via Pythagorean identities. Of course, the Pythagorean theorem is best known by the equation a^2 + b^2 = c^2. To extend this to trigonometry, let (x, y) be an ordered pair on the unit circle, i.e. the circle centered at the origin and with radius equal to 1. By our famous theorem, we have that x^2 + y ^2 = 1, since the coordinates x and y form a right triangle with hypotenuse 1. It is from this construct that we obtain the trigonometric identities, which we explore here.

Let us recall the definitions of the sine and cosine functions on the unit circle with equation x^2 + y^2 = 1. To understand this, it is important to know that the abscissa is the abscissa and the y- coordinate is the ordinate. With this in mind, we define the sine as the ordinate/radius and the cosine as the abscissa/radius. By denoting respectively x and y as abscissa and ordinate, and r as radius, and A as generated angle, we have sin(A) = y/r and cos(A) = x/r.

Since r = 1, sin(A) = y and cos(A) = x in the previous definitions. Since we know that x^2 + y^2 = 1, we have sin^2(A) + cos^2(A) = 1. This is our first Pythagorean identity based on the unit circle. Now there are two more based on the other trigonometric functions, namely tangent, cotangent, secant and cosecant. Luckily, we only need to memorize the first one because the other two are free, as my calculus teacher learned during my freshman year in college. The way to derive the other two identities is based on the relationship between tangent (tan) and cotangent (cot); and secant (sec) and cosecant (csc).

Reciprocal identities

To derive the other two Pythagorean identities, we use the reciprocal identities below:

csc(A) = 1/sin(A)

sec(A) = 1/cos(A)

lit(A) = 1/tan(A)

tan(A) = sin(A)/cos(A)

As my college calculus teacher demonstrated to me, we start with the first one and successively derive the others as follows:

(1) sin^2(A) + cos^2(A) = 1

To obtain the Pythagorean identity involving tan and cot, we divide the entire equation by cos^2(A). This gives

sin^2(A)/cos^2(A) + cos^2(A)/cos^2(A) = 1/cos^2(A)

Using the reciprocal identities above, we see that this equation is the same as

tan^2(A) + 1 = sec^2(A)

To obtain the Pythagorean identity involving cot and csc, we divide equation (1) above by sin^2(A), again resorting to our reciprocal identities to obtain

sin^2(A)/sin^2(A) + cos^2(A)/sin^2(A) = 1/sin^2(A)

Simplifying, this gives our third Pythagorean identity:

1 + cot^2(A) = csc^2(A)

That’s really all there is to it. And so it is my dear friends that we use one identity to get two more for free. Maybe there are no free lunches in life, but at least sometimes there are free lunches in math. Thank God!

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