The famous Pythagorean Theorem extends over to trigonometry via the Pythagorean identities. Of course, the Pythagorean Theorem is most remembered by the equation a^2 + b^2 = c^2. To extend this to trigonometry, we let (x, y) be an ordered pair on the unit circle, that is the circle centered at the origin and having radius equal to 1. By our famous theorem, we have that x^2 + y^2 = 1, since the x and y coordinates carve out a right triangle of 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 of equation x^2 + y^2 = 1. In order to understand this, it is important to know that the x-coordinate 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. Denoting x and y as the abscissa and ordinate, respectively, and r as the radius, and A as the angle generated, 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 others based on the other trigonometric functions, namely the tangent, cotangent, secant, and cosecant. Fortunately though we need only memorize the first one because the other two come free, as I was taught by my Calculus I professor 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).
To derive the other two Pythagorean identities, we use the reciprocal identities below:
csc(A) = 1/sin(A)
sec(A) = 1/cos(A)
cot(A) = 1/tan(A)
tan(A) = sin(A)/cos(A)
As my college calculus professor 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 get 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 get 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)
Upon simplifying, this gives our third Pythagorean identity:
1 + cot^2(A) = csc^2(A)
That’s really all there is to it. And that my dear friends is how we use one identity to obtain two others for free. Maybe there are no free lunches in life, but at least sometimes there are free lunches in mathematics. Thank God!