The Pythagorean identities

When you reach for it

The reciprocal and quotient identities write tangent, cotangent, secant and cosecant in terms of sine and cosine. They say nothing about how sine and cosine relate to each other. Yet the two are not free: they are the coordinates of one point, and that point lies on a circle. The equation of that circle, written in the new names, is the Pythagorean identity, and dividing it through gives two more.

Reach for these identities whenever one trigonometric value of an angle is known and another is wanted, especially when the angle itself is never found. Reach for them also whenever an expression contains a square such as \sin^2\theta or \sec^2\theta, since each identity lets one square be traded for another, and whenever you need to test whether two numbers could be the sine and cosine of the same angle.

The procedure

Where \sin^2\theta + \cos^2\theta = 1 comes from

Let the terminal side of \theta meet the unit circle at (x, y). Every point of that circle is at distance 1 from the origin, so by the distance formula, which is the Pythagorean theorem applied to the coordinates,

On the unit circle x = \cos\theta and y = \sin\theta. Renaming the coordinates gives (\cos\theta)^2 + (\sin\theta)^2 = 1. The square of a sine is written \sin^2\theta, with the exponent after the name, so that it is not mistaken for \sin(\theta^2):

In Quadrant I this is the Pythagorean theorem for the triangle in the figure, with legs \cos\theta and \sin\theta and hypotenuse 1. In the other quadrants a coordinate is negative, but squaring removes the sign, and on an axis the triangle flattens while the equation still holds: at \theta = \pi it reads 0^2 + (-1)^2 = 1. Sine and cosine are defined at every real number, so the identity has no exceptions. It is often used rearranged, as \sin^2\theta = 1 - \cos^2\theta or \cos^2\theta = 1 - \sin^2\theta.

Dividing by \cos^2\theta and by \sin^2\theta

At any angle where \cos\theta \neq 0, divide both sides by \cos^2\theta:

The first fraction is \displaystyle \left(\frac{\sin\theta}{\cos\theta}\right)^2 = \tan^2\theta by the quotient identity, the second is 1, and the right side is \sec^2\theta by the reciprocal identity. So \tan^2\theta + 1 = \sec^2\theta. The division was allowed only where \cos\theta \neq 0, and those are exactly the angles where tangent and secant exist, so nothing is lost: the identity holds on its whole common domain, \displaystyle \theta \neq \frac{\pi}{2} + k\pi for every integer k.

Dividing instead by \sin^2\theta, where \sin\theta \neq 0, turns the three terms into 1, \cot^2\theta and \csc^2\theta, so 1 + \cot^2\theta = \csc^2\theta for \theta \neq k\pi.

The Pythagorean identities. For every integer k:

From one value and a sign to the other five

Each identity links a pair: sine with cosine, tangent with secant, cotangent with cosecant. Given one value, pick the identity that contains it and its partner, and solve for the partner’s square. A square root then gives the partner’s size but not its sign, because two numbers have each positive square. The sign comes from the quadrant, which a second condition such as “\sin\theta > 0” decides, exactly as in finding a quadrant from two signs. Once two values are known, the reciprocal and quotient identities give the other four without any more square roots. For instance, multiplying \displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta} by the nonzero \cos\theta gives \sin\theta = \tan\theta\cos\theta.

That earlier lesson found the same six values by drawing a reference triangle, or a point on the terminal ray, and using its sides. The two methods always agree, because they are one equation. A right triangle with legs a, b and hypotenuse c has a^2 + b^2 = c^2; dividing by c^2 gives \displaystyle \left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1, which is the Pythagorean identity for the triangle shrunk to hypotenuse 1. Example 2 works one problem both ways. The identity saves drawing, and it still works when the given value is a letter rather than a number (Example 5).

Worked examples

Boundaries

The tangent and cotangent forms have gaps. 1 + \tan^2\theta = \sec^2\theta says nothing at \displaystyle \theta = \frac{\pi}{2} + k\pi, and 1 + \cot^2\theta = \csc^2\theta nothing at \theta = k\pi. Only \sin^2\theta + \cos^2\theta = 1 holds at every angle.

A square root gives a size, not a sign. \displaystyle \sqrt{1 - \cos^2\theta} = |\sin\theta|, not \sin\theta. The sign must come from the quadrant, and a given value with no second condition leaves two answers. When a value puts the terminal side on an axis, there is no quadrant, and some of the six may not exist: \sin\theta = 1 forces \cos\theta = 0, so tangent and secant are undefined.

Not every number is possible. Since squares are not negative, the identity forces -1 \le \sin\theta \le 1 and -1 \le \cos\theta \le 1, and then \sec^2\theta \ge 1 and \csc^2\theta \ge 1. A value such as \displaystyle \sec\theta = \frac{1}{2} belongs to no angle: it would need \displaystyle \tan^2\theta = -\frac{3}{4}. Tangent and cotangent may be any real number.

One angle throughout. \sin^2 A + \cos^2 B is not 1 unless the angles agree in the right way, and \sin^2\theta is (\sin\theta)^2, not \sin(\theta^2).

Where this goes next

With the reciprocal, quotient and Pythagorean identities in hand, the next move is to use them to rewrite: simplifying expressions trades 1 - \sin^2\theta for \cos^2\theta and \sec^2\theta - 1 for \tan^2\theta, and verifying identities proves new statements from these three. In trigonometric equations the identity turns an equation mixing \sin^2 x and \cos x into a quadratic in \cos x alone. The distance between two points on the unit circle, computed with \sin^2 + \cos^2 = 1, is how the cosine of a sum is derived, and the bound |\sin\theta| \le 1 that the identity forces is one ingredient of the squeeze argument for \displaystyle \frac{\sin x}{x}.

Exercises

Values the identities give at once

Find each exact value.

1. 7\sin^2 20^\circ + 7\cos^2 20^\circ
2. \displaystyle \tan^2\frac{\pi}{9} - \sec^2\frac{\pi}{9}
3. \cot^2 2 - \csc^2 2, with 2 in radians

Can these belong to one angle?

4. Is there an angle with \sin\theta = 0.6 and \cos\theta = -0.7?
5. Is there an angle with \tan\theta = -4 and \displaystyle \sec\theta = \sqrt{17}? If so, name its quadrant.
6. Find every number k for which some angle has \sin\theta = k and \cos\theta = 2k.

All six from one value and a sign

Find the other five values exactly, with rational denominators.

7. \displaystyle \cos\theta = \frac{2}{7} and \sin\theta < 0
8. \displaystyle \sin\theta = \frac{39}{89} and \cos\theta < 0
9. \displaystyle \tan\theta = \frac{\sqrt{7}}{3} and \sin\theta < 0
10. \csc\theta = -6 and \tan\theta < 0
11. \displaystyle \sec\theta = \frac{9}{4} and \sin\theta < 0
12. \displaystyle \cot\theta = \frac{1}{2} and \csc\theta < 0

One function in terms of another

13. \theta is in Quadrant IV. Write \cos\theta and \tan\theta in terms of s = \sin\theta.
14. \theta is in Quadrant I. Write \cos\theta and \sin\theta in terms of t = \tan\theta.
15. \theta is in Quadrant III. Write \sin\theta and \tan\theta in terms of c = \cos\theta.

Sums and products

16. \displaystyle \sin\theta + \cos\theta = \frac{17}{13} and \sin\theta > \cos\theta. Find \sin\theta and \cos\theta.
17. \displaystyle \sin\theta - \cos\theta = \frac{1}{3}. Find \sin\theta\cos\theta.

Where a square root form holds

For which \theta with 0 \le \theta < 2\pi is each equation true?

18. \displaystyle \cos\theta = -\sqrt{1 - \sin^2\theta}
19. \displaystyle \csc\theta = \sqrt{1 + \cot^2\theta}

Reasoning

20. Explain why \sin^2\theta + \cos^2\theta = 1 holds at \displaystyle \theta = \frac{\pi}{2} but 1 + \tan^2\theta = \sec^2\theta says nothing there.
21. Show by one angle that \displaystyle \sin\theta = \sqrt{1 - \cos^2\theta} is not an identity, and say for which \theta it is true.
22. Derive 1 + \cot^2\theta = \csc^2\theta directly from x^2 + y^2 = 1, saying where the step is allowed.
23. Explain why the Pythagorean identities force |\cos\theta| \le 1 and |\sec\theta| \ge 1.
24. In Example 1, a reference triangle with hypotenuse 7 and adjacent side 3 gives the same values. Explain, with the equation for that triangle, why it must.