Introduction
Sooner or later, learning trigonometry, you meet this equation.
$$\sin^{2}\theta+\cos^{2}\theta=1$$Here \(\sin^{2}\theta\) means 'the sine value squared' and \(\cos^{2}\theta\) means 'the cosine value squared' (\(\theta\) is theta, the Greek letter for an angle). The striking thing is that whether you set the angle \(\theta\) to \(0°\), \(37°\), or \(200°\), the sum of the two squares is always exactly 1. Sine and cosine rise and fall endlessly with the angle, yet square each and add them and all that wobbling cancels perfectly, pinned to 1.
At first sight it looks like a magic formula to memorize whole. But this equation is nothing more than a theorem you already know, rewritten in the language of trigonometry — the Pythagorean theorem. This post confirms that with your own eyes (Section 1), notes that the tangent is really just the ratio of sine to cosine (Section 2), and then shows how dividing that single line just once produces another identity linking tangent and secant on its own (Section 3). You will see that there is not a pile of formulas to memorize, but really one equation, repackaged into a family.
1. sin²θ+cos²θ=1 is really the Pythagorean theorem
First move the stage onto the unit circle — the circle with center at the origin and radius \(1\). As an earlier post showed, if \(P\) is the point where the terminal side rotated by \(\theta\) (a ray reaching out from the origin) meets this circle, then that point's coordinates are exactly
$$P=(\cos\theta,\ \sin\theta)$$the horizontal coordinate being the cosine and the vertical the sine. If you want to know why, the unit circle post in 'Related reading' below sets it up; here we take that result as our starting point.
Now we need ask only one key question. How far is the point \(P\) from the origin? Since the unit circle has radius \(1\), by definition \(P\) sits exactly \(1\) away from the origin. On the other hand, in the coordinate plane the distance between the origin \((0,0)\) and a point \((x,y)\) is, by the Pythagorean theorem,
$$\text{distance}=\sqrt{x^{2}+y^{2}}$$because that distance is the hypotenuse of the right triangle whose legs run \(x\) across and \(y\) up. Plug in \(P\)'s coordinates \(x=\cos\theta\), \(y=\sin\theta\), and use the fact that the distance is \(1\):
$$\sqrt{\cos^{2}\theta+\sin^{2}\theta}=1$$and squaring both sides to strip the root gives
$$\cos^{2}\theta+\sin^{2}\theta=1$$That is all there is to it. The square of the sine plus the square of the cosine is just the sum of the squared legs of the right triangle drawn from the point on the unit circle, and that value is the square of the hypotenuse — the radius \(1\) — namely \(1\). However much the angle changes, the point stays on the same circle, so its distance is always \(1\), and the sum of the two squares is nailed at \(1\). Not a new formula to memorize, but one line rewriting the Pythagorean theorem in sines and cosines.
Let us handle it with numbers. At \(\theta=60°\) the cosine is \(0.5\) and the sine is \(0.866\). Squaring each,
$$\cos^{2}60°=0.5^{2}=0.25,\qquad \sin^{2}60°=0.866^{2}=0.75$$and adding gives \(0.25+0.75=1\). The two values themselves differ — \(0.25\) and \(0.75\) — but the sum is unfailingly \(1\). Change the angle and only this split between \(0.25\) and \(0.75\) shifts; the total stays put.
2. The tangent is just the ratio of sine to cosine
Before the third identity, let us settle what the tangent is. From the coordinate definition on the unit circle,
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$the vertical coordinate (sine) divided by the horizontal (cosine). This is no new convention: it is the right triangle's 'height over base' tangent, rewritten with height \(=\sin\theta\) and base \(=\cos\theta\). In short, the tangent carries no information beyond sine and cosine. Hold sine and cosine and the tangent follows from a single division.
One more function to introduce: the secant. The secant (\(\sec\) for short) is the reciprocal of the cosine — \(1\) divided by the cosine.
$$\sec\theta=\frac{1}{\cos\theta}$$Only the name is unfamiliar; there is nothing grand about it. If the cosine is \(0.5\) the secant is \(2\); if the cosine is \(0.8\) the secant is \(1.25\). With these two supporting players in hand — the tangent as the ratio of sine to cosine, the secant as the reciprocal of cosine — the new identity in the next section falls out from a single calculation.
3. 1+tan²θ=sec²θ — the same equation, divided by cos²
Here is the last step, the one that looks like a trick. Take the identity from Section 1 again.
$$\sin^{2}\theta+\cos^{2}\theta=1$$Divide both sides by \(\cos^{2}\theta\). An equation still holds when you divide both sides by the same thing (provided \(\cos\theta\ne0\), that is, as long as what you divide by is not \(0\)). The left side has two terms, so divide each:
$$\frac{\sin^{2}\theta}{\cos^{2}\theta}+\frac{\cos^{2}\theta}{\cos^{2}\theta}=\frac{1}{\cos^{2}\theta}$$Now read the three pieces one by one.
- The first term \(\dfrac{\sin^{2}\theta}{\cos^{2}\theta}\) is \(\left(\dfrac{\sin\theta}{\cos\theta}\right)^{2}\), namely \(\tan^{2}\theta\) (the tangent definition from Section 2, squared).
- The second term \(\dfrac{\cos^{2}\theta}{\cos^{2}\theta}\) is a thing divided by itself, so simply \(1\).
- The right side \(\dfrac{1}{\cos^{2}\theta}\) is \(\left(\dfrac{1}{\cos\theta}\right)^{2}\), namely \(\sec^{2}\theta\) (the secant definition, squared).
Reassemble the three pieces:
$$\tan^{2}\theta+1=\sec^{2}\theta$$a second identity, complete. Nothing new to prove, nothing new to memorize. We only divided the one line from Section 1 by \(\cos^{2}\theta\) and reattached the labels. In the same way, dividing instead by \(\sin^{2}\theta\) makes the cousin identity \(1+\cot^{2}\theta=\csc^{2}\theta\), linking cotangent and cosecant, pop out by the same route. The identities looked like several, but the root was a single Pythagorean theorem.
Check it with numbers. Taking \(\theta=60°\) as before, the tangent is \(1.732\) and the secant, the reciprocal of the cosine \(0.5\), is \(2\). Then
$$\tan^{2}60°=1.732^{2}=3,\qquad \sec^{2}60°=2^{2}=4$$and the left side is \(\tan^{2}60°+1=3+1=4\), exactly matching the right side \(\sec^{2}60°=4\). See for yourself that this \(3+1=4\) is just the \(0.25+0.75=1\) of Section 1 divided by \(\cos^{2}60°=0.25\): \(0.75/0.25=3\) (\(=\tan^{2}\)), \(0.25/0.25=1\), and \(1/0.25=4\) (\(=\sec^{2}\)). Two faces of the same equation.
4. See it for yourself
In the interactive below, sweep the \(\theta\) slider. On the left, the point \(P\) moves around the unit circle; on the right, the bar stacks \(\cos^{2}\theta\) (blue) and \(\sin^{2}\theta\) (green) one atop the other. However the angle changes, the two bars together always reach exactly to the mark at \(1\) — that is the first identity. Only the split between the two pieces shifts; the total height stays fixed.
Press the '÷ cos²θ' button and the same equation switches to its divided-by-cosine-squared form. Now the bar stacks \(1\) (blue) and \(\tan^{2}\theta\) (green), and their sum reaches the mark at \(\sec^{2}\theta\) — the second identity. The 'example angles' buttons for \(30°\), \(45°\), and \(60°\) reproduce the values computed in the text (\(0.25+0.75=1\) and \(3+1=4\) at \(60°\)).
Key takeaways
- \(\sin^{2}\theta+\cos^{2}\theta=1\) is not a magic formula to memorize but one line rewriting the Pythagorean theorem in trigonometric functions. The point \((\cos\theta,\sin\theta)\) on the unit circle is always a distance of \(1\) from the origin, and writing that distance with Pythagoras makes the sum of the squared coordinates equal the squared radius, \(1\).
- The tangent is \(\tan\theta=\sin\theta/\cos\theta\), merely the ratio of sine to cosine, and the secant is the reciprocal of the cosine, \(\sec\theta=1/\cos\theta\). Neither carries information beyond sine and cosine.
- Divide that one identity by \(\cos^{2}\theta\) and reattach the labels, and \(1+\tan^{2}\theta=\sec^{2}\theta\) comes out on its own. Divide by \(\sin^{2}\theta\) and you get \(1+\cot^{2}\theta=\csc^{2}\theta\). The whole family of trigonometric identities is a single Pythagorean theorem, repackaged.
Once you see that what looked like several formulas is really variations on one, the burden of memorizing drops away. And this identity is the most basic tool for handling trigonometric functions — swapping sine for cosine (or the reverse) to tidy an expression. It is the first wrench you reach for when computing an integral or combining waves.
Related reading — A Right Triangle Only Reaches 90° — So How Do We Measure sin 120°? (the definition where a point's coordinates on the unit circle become sine and cosine — this post's starting point) · Why Are Trigonometric Ratios the Same Regardless of a Triangle's Size (the root of the tangent as a ratio of sides) · What Is the Point of Trigonometric Functions Anyway (where these identities lead on into combining waves)