Cofunction identities
The cosine of an acute angle is the sine of its complement. Does that swap survive for every real number t, where no right triangle can be drawn?
Where you are 13 in Trigonometric identities
Why this exists
In a right triangle the two acute angles add to 90^\circ, and the side opposite one is the side adjacent to the other. Sine, cosine and tangent read that as \sin A = \cos(90^\circ - A), the “co” in cosine standing for complement, and Secant, cosecant and cotangent ratios did the same for secant with cosecant and tangent with cotangent.
That argument needs a triangle, so it only covers angles between 0 and \displaystyle \frac{\pi}{2}. On the unit circle every real number t has a sine and a cosine, and t = 2.5 or t = -1 is no angle of any right triangle. Is \displaystyle \sin\left(\frac{\pi}{2} - 2.5\right) still \cos 2.5? It is, and the reason is a fourth mirror of the unit circle, the line y = x, alongside the three in Unit circle symmetry and even-odd identities. Once the swap holds for every t, it combines with the even and odd identities and with periodicity to turn any quarter-turn shift, such as \displaystyle \sin\left(t - \frac{\pi}{2}\right) or \displaystyle \cos\left(\frac{\pi}{2} + t\right), into a single function of t.
Definition
In radians, two numbers are complementary when they add to \displaystyle \frac{\pi}{2}, so the complement of t is \displaystyle \frac{\pi}{2} - t. Unlike the angles of a triangle, t may be any real number, and its complement may be negative or larger than \displaystyle \frac{\pi}{2}.
The six trig functions come in three cofunction pairs: sine and cosine, tangent and cotangent, secant and cosecant. The cofunction identities say that each function of t equals its partner function of the complement of t, for every real t at which the two sides are defined. They are identities in the sense of What an identity is. In degrees the complement of \theta is 90^\circ - \theta.
Intuition
The unit circle is symmetric about the line y = x: replacing (x, y) by (y, x) turns x^2 + y^2 = 1 into y^2 + x^2 = 1, the same equation. So swapping the coordinates of a point on the circle gives another point on the circle.
That mirror swaps the two places where angles are measured from. It sends (1, 0), where every angle starts, to (0, 1), the terminal point of \displaystyle \frac{\pi}{2}. It also reverses the direction of turning, as every reflection does. So the walk from (1, 0) through t counterclockwise becomes a walk from (0, 1) through t clockwise, and it ends at the point \displaystyle \frac{\pi}{2} - t along the circle. In the figure t is about 125^\circ, and the walk back from the top of the circle overshoots (1, 0) and ends in Quadrant IV at -35^\circ; nothing about the swap needed t to be acute.
Results
Sine and cosine. For every real number t,
The other two pairs. For every real t at which both sides are defined,
and each side is defined exactly when the other is. The identities for \displaystyle \tan\left(\frac{\pi}{2} - t\right) and \displaystyle \sec\left(\frac{\pi}{2} - t\right) need \sin t \ne 0, so t \ne k\pi; those for \displaystyle \cot\left(\frac{\pi}{2} - t\right) and \displaystyle \csc\left(\frac{\pi}{2} - t\right) need \cos t \ne 0, so \displaystyle t \ne \frac{\pi}{2} + k\pi; here k is any integer.
Quarter-turn shifts. Combined with the even and odd identities, for every real t,
So adding a quarter turn sends the point (x, y) to (-y, x): the rotation by \displaystyle \frac{\pi}{2}.
Why it is true
The swap. Let P = (\cos t, \sin t) be the terminal point of t, reached from (1, 0) by an arc of length |t|, counterclockwise if t > 0 and clockwise if t < 0, wrapping around as many times as it needs. The reflection (x, y) \to (y, x) keeps the circle, keeps arc lengths, sends (1, 0) to (0, 1) and reverses direction. So it carries that arc to an arc of the same length from (0, 1), turning the other way. The point (0, 1) sits at \displaystyle \frac{\pi}{2}, so the image arc ends at \displaystyle \frac{\pi}{2} - t. The image of P is (\sin t, \cos t), and it is the terminal point of \displaystyle \frac{\pi}{2} - t:
Reading the two coordinates gives both sine and cosine identities. No step used \displaystyle 0 < t < \frac{\pi}{2}: a negative t or a t beyond a full turn gives a longer arc or one run the other way, and the mirror handles it the same way.
The other four. Each is built from sine and cosine (see Reciprocal and quotient identities), so substitution does the work:
and in the same way \displaystyle \cot\left(\frac{\pi}{2} - t\right) = \frac{\sin t}{\cos t} = \tan t and \displaystyle \csc\left(\frac{\pi}{2} - t\right) = \frac{1}{\cos t} = \sec t. Each left side divides by \sin t or \cos t, exactly as its right side does, so the two sides are undefined at the same t.
The shifts. Write \displaystyle t + \frac{\pi}{2} as \displaystyle \frac{\pi}{2} - (-t) and use the swap, then the even and odd identities:
Worked examples
Rewriting a value as its cofunction
Write each as the cofunction of another angle: (a) \displaystyle \sin\frac{2\pi}{7}; (b) \displaystyle \cot\frac{5\pi}{8}; (c) \csc(-0.4).
SolutionThe partner function takes the complement \displaystyle \frac{\pi}{2} - t, worked over a common denominator.
(a) \displaystyle \frac{\pi}{2} - \frac{2\pi}{7} = \frac{7\pi}{14} - \frac{4\pi}{14} = \frac{3\pi}{14}, so \displaystyle \sin\frac{2\pi}{7} = \cos\frac{3\pi}{14}.
(b) \displaystyle \frac{\pi}{2} - \frac{5\pi}{8} = \frac{4\pi}{8} - \frac{5\pi}{8} = -\frac{\pi}{8}, so \displaystyle \cot\frac{5\pi}{8} = \tan\left(-\frac{\pi}{8}\right). The complement is negative, which the triangle version could not allow and the unit circle version does. Both values are negative: \displaystyle \frac{5\pi}{8} is in Quadrant II, where cotangent is negative, and \displaystyle -\frac{\pi}{8} is in Quadrant IV, where tangent is negative.
(c) \displaystyle \frac{\pi}{2} - (-0.4) = \frac{\pi}{2} + 0.4, so \displaystyle \csc(-0.4) = \sec\left(\frac{\pi}{2} + 0.4\right). Check on a calculator: both are about -2.568.
All six functions of the complement
The terminal point of t is \displaystyle \left(-\frac{51}{149}, -\frac{140}{149}\right). Find the terminal point of \displaystyle \frac{\pi}{2} - t and all six trig functions of \displaystyle \frac{\pi}{2} - t.
SolutionThe point is on the unit circle, since 51^2 + 140^2 = 2{,}601 + 19{,}600 = 22{,}201 = 149^2. The terminal point of \displaystyle \frac{\pi}{2} - t is its mirror image across y = x, the same coordinates swapped:
Both points are in Quadrant III, because swapping two negative numbers leaves two negative numbers: that mirror carries Quadrant III to itself. Reading the coordinates,
Check against the identities: \displaystyle \cot t = \frac{\cos t}{\sin t} = \frac{51}{140}, matching \displaystyle \tan\left(\frac{\pi}{2} - t\right), and \displaystyle \csc t = -\frac{149}{140}, matching \displaystyle \sec\left(\frac{\pi}{2} - t\right).
Shifts by a quarter turn the other way
Simplify (a) \displaystyle \sin\left(t - \frac{\pi}{2}\right) and (b) \displaystyle \cos\left(\frac{\pi}{2} + t\right) to a single function of t.
Solution(a) The angle \displaystyle t - \frac{\pi}{2} is the opposite of \displaystyle \frac{\pi}{2} - t. Sine is odd, so
(b) Write \displaystyle \frac{\pi}{2} + t as \displaystyle \frac{\pi}{2} - (-t). The cofunction identity and then oddness of sine give
Check (a) at t = 0: \displaystyle \sin\left(-\frac{\pi}{2}\right) = -1 = -\cos 0. Check (b) at \displaystyle t = \frac{\pi}{2}: \displaystyle \cos\pi = -1 = -\sin\frac{\pi}{2}. On the graphs, (a) says that sliding the sine curve \displaystyle \frac{\pi}{2} to the right gives the cosine curve turned upside down (Graphing sine and cosine).
Larger shifts, with periodicity
Simplify (a) \displaystyle \cos\left(\frac{3\pi}{2} - t\right); (b) \displaystyle \tan\left(t + \frac{\pi}{2}\right); (c) \displaystyle \sin\left(t + \frac{5\pi}{2}\right).
Solution(a) \displaystyle \frac{3\pi}{2} - t = \pi + \left(\frac{\pi}{2} - t\right). The half turn changes the sign of cosine, \cos(\pi + u) = -\cos u from Unit circle symmetry and even-odd identities, so
(b) \displaystyle t + \frac{\pi}{2} = \frac{\pi}{2} - (-t), so the cofunction identity and oddness of cotangent give \displaystyle \tan\left(t + \frac{\pi}{2}\right) = \cot(-t) = -\cot t. Both sides are undefined exactly when \sin t = 0, that is at t = k\pi for every integer k.
(c) \displaystyle \frac{5\pi}{2} = \frac{\pi}{2} + 2\pi, and sine repeats every 2\pi (Periodicity and the periodicity identities), so
Check (a) at \displaystyle t = \frac{\pi}{2}: \displaystyle \cos\pi = -1 = -\sin\frac{\pi}{2}.
Simplifying products
Simplify, and say for which t the result holds: (a) \displaystyle \cos\left(\frac{\pi}{2} - t\right)\csc\left(\frac{\pi}{2} - t\right); (b) \displaystyle \tan\left(\frac{\pi}{2} - t\right)\tan t.
Solution(a) Replace each factor by its cofunction: \displaystyle \cos\left(\frac{\pi}{2} - t\right) = \sin t and \displaystyle \csc\left(\frac{\pi}{2} - t\right) = \sec t. Then
The original expression needs \displaystyle \csc\left(\frac{\pi}{2} - t\right) defined, that is \cos t \ne 0, which is exactly where \tan t is defined.
(b) \displaystyle \tan\left(\frac{\pi}{2} - t\right) = \cot t, and \cot t \tan t = 1 because the two are reciprocals. The original needs both \tan t (so \cos t \ne 0) and \cot t (so \sin t \ne 0), so the product is 1 for every t except the integer multiples of \displaystyle \frac{\pi}{2}, where it is undefined.
Veikko’s two angles
Veikko knows that \sin t = \cos 0.4. Find every t in [0, 2\pi) that fits, to three decimal places.
SolutionThe cofunction identity turns the cosine into a sine: \displaystyle \cos 0.4 = \sin\left(\frac{\pi}{2} - 0.4\right), so one answer is
The sine is the y-coordinate, and the horizontal line at that height meets the circle twice: at this point and at its mirror image across the y-axis, the terminal point of \pi - t (Unit circle symmetry and even-odd identities). That gives
Both are in [0, 2\pi), and the line meets the circle in only two points, so there are no others. Check: \sin 1.971 \approx 0.921 \approx \cos 0.4. The two answers sit symmetrically about \displaystyle \frac{\pi}{2}, where the sine curve peaks.
Boundaries
- The tangent, cotangent, secant and cosecant identities hold only where defined. At t = 0, \displaystyle \tan\left(\frac{\pi}{2} - t\right) = \cot t says nothing, since both sides are undefined.
- The swap is for the complement \displaystyle \frac{\pi}{2} - t exactly. For \displaystyle t - \frac{\pi}{2} or \displaystyle t + \frac{\pi}{2} a sign appears, as in Example 3, because those are a quarter turn of the circle, not a reflection.
- The units must match: \displaystyle \frac{\pi}{2} - t in radians, 90^\circ - \theta in degrees. \sin(90 - 0.3), with 0.3 in radians, is not \cos 0.3.
- A function of the complement equals the partner function, never the same function: \displaystyle \sin\left(\frac{\pi}{2} - t\right) is \cos t, and it equals \sin t only when \sin t = \cos t, at \displaystyle t = \frac{\pi}{4} + k\pi.
Where this goes next
Sine and tangent of a sum or difference gets the sine of a sum from the cosine of a difference (Cosine of a sum or difference) by exactly this swap: \displaystyle \sin(a + b) = \cos\left(\frac{\pi}{2} - a - b\right). Once the cosine difference formula is proved there, it will also give a second proof of the cofunction identities, with no figure. The quarter-turn shifts say the sine and cosine graphs are one curve slid by \displaystyle \frac{\pi}{2}, which is why a sinusoid can be written with either function and a phase shift (Phase shift). In Simplifying trig expressions and Verifying identities the cofunction identities sit beside the quotient, reciprocal and Pythagorean identities as the basic tools.
Exercises
Writing a value as its cofunction
- Write \displaystyle \sin\frac{3\pi}{10} as the cosine of another angle.
- Write \displaystyle \cot\frac{\pi}{12} as the tangent of another angle.
- Write \displaystyle \sec\frac{5\pi}{7} as the cosecant of another angle.
- Write \cos(-1.3) as the sine of another angle.
- Write \csc 128^\circ as the secant of another angle, in degrees.
Terminal points and values
- The terminal point of t is \displaystyle \left(-\frac{88}{137}, \frac{105}{137}\right). Find the terminal point of \displaystyle \frac{\pi}{2} - t, and find \displaystyle \tan\left(\frac{\pi}{2} - t\right) and \displaystyle \sec\left(\frac{\pi}{2} - t\right).
- An angle t has \cos t = 0.6 and \sin t = -0.8. Find all six trig functions of \displaystyle \frac{\pi}{2} - t.
- Given \sin 0.3 \approx 0.2955 and \cos 0.3 \approx 0.9553, find \displaystyle \sin\left(\frac{\pi}{2} - 0.3\right), \displaystyle \cos\left(\frac{\pi}{2} + 0.3\right) and \displaystyle \sin\left(0.3 - \frac{\pi}{2}\right).
Quarter turns and simplifying
- Simplify \displaystyle \cos\left(t - \frac{\pi}{2}\right).
- Simplify \displaystyle \sin\left(\frac{\pi}{2} + t\right).
- Simplify \displaystyle \cot\left(t + \frac{\pi}{2}\right).
- Simplify \displaystyle \sec\left(t - \frac{\pi}{2}\right).
- Simplify \displaystyle \sin\left(\frac{3\pi}{2} - t\right).
- Simplify \displaystyle \cos\left(t + \frac{3\pi}{2}\right).
- Simplify \displaystyle \sin t\cos\left(\frac{\pi}{2} - t\right) + \cos t\sin\left(\frac{\pi}{2} - t\right).
- Simplify \displaystyle \dfrac{\sec\left(\frac{\pi}{2} - t\right)}{\csc\left(\frac{\pi}{2} - t\right)} and say for which t the result holds.
Finding angles
- Find every t in [0, 2\pi) with \cos t = \sin 0.7, to three decimal places.
- Toivo needs every t in [0, 2\pi) with \displaystyle \tan t = \cot\frac{\pi}{5}. Find them exactly.
- Find every \theta in [0^\circ, 360^\circ) with \sin\theta = \cos 200^\circ.
Reasoning
- Explain, with arcs, why reflecting across y = x sends the terminal point of t = -0.5 to the terminal point of \displaystyle \frac{\pi}{2} + 0.5.
- Hilja proved \displaystyle \sin\left(\frac{\pi}{2} - t\right) = \cos t with a right triangle. Explain which values of t her proof covers, and what the unit circle argument adds.
- Prove \displaystyle \csc\left(\frac{\pi}{2} - t\right) = \sec t from the sine and cosine identities, and say for which t it holds.
- Veikko says \displaystyle \cos\left(\frac{\pi}{2} - t\right) = \sin t follows from \displaystyle \sin\left(\frac{\pi}{2} - t\right) = \cos t with no figure at all. Show how, by substituting a new letter for \displaystyle \frac{\pi}{2} - t.
- If the terminal point of t is in Quadrant II, in which quadrant is the terminal point of \displaystyle \frac{\pi}{2} - t? Explain with the swap, and check with the signs of \displaystyle \sin\left(\frac{\pi}{2} - t\right) and \cos t.
Answers
- \displaystyle \cos\frac{\pi}{5}. \displaystyle \frac{\pi}{2} - \frac{3\pi}{10} = \frac{5\pi}{10} - \frac{3\pi}{10} = \frac{2\pi}{10}.
- \displaystyle \tan\frac{5\pi}{12}. \displaystyle \frac{6\pi}{12} - \frac{\pi}{12} = \frac{5\pi}{12}.
- \displaystyle \csc\left(-\frac{3\pi}{14}\right). \displaystyle \frac{7\pi}{14} - \frac{10\pi}{14} = -\frac{3\pi}{14}.
- \displaystyle \sin\left(\frac{\pi}{2} + 1.3\right). The complement of -1.3 is \displaystyle \frac{\pi}{2} - (-1.3).
- \sec(-38^\circ). 90^\circ - 128^\circ = -38^\circ.
- Terminal point \displaystyle \left(\frac{105}{137}, -\frac{88}{137}\right), the coordinates swapped. \displaystyle \tan\left(\frac{\pi}{2} - t\right) = -\frac{88}{105}, which is \cot t; \displaystyle \sec\left(\frac{\pi}{2} - t\right) = \frac{137}{105}, which is \csc t.
- \displaystyle \cos\left(\frac{\pi}{2} - t\right) = -0.8, \displaystyle \sin\left(\frac{\pi}{2} - t\right) = 0.6, \displaystyle \tan\left(\frac{\pi}{2} - t\right) = -0.75, \displaystyle \cot\left(\frac{\pi}{2} - t\right) = -\frac{4}{3}, \displaystyle \sec\left(\frac{\pi}{2} - t\right) = -1.25, \displaystyle \csc\left(\frac{\pi}{2} - t\right) = \frac{5}{3}. The first two are \sin t and \cos t; the rest come from them.
- \displaystyle \sin\left(\frac{\pi}{2} - 0.3\right) \approx 0.9553, which is \cos 0.3; \displaystyle \cos\left(\frac{\pi}{2} + 0.3\right) \approx -0.2955, which is -\sin 0.3; \displaystyle \sin\left(0.3 - \frac{\pi}{2}\right) \approx -0.9553, which is -\cos 0.3.
- \sin t. Cosine is even, so it equals \displaystyle \cos\left(\frac{\pi}{2} - t\right) = \sin t.
- \cos t. \displaystyle \frac{\pi}{2} + t = \frac{\pi}{2} - (-t), so it is \cos(-t) = \cos t.
- -\tan t. It is \displaystyle \cot\left(\frac{\pi}{2} - (-t)\right) = \tan(-t) = -\tan t, wherever \cos t \ne 0.
- \csc t. Secant is even, so it is \displaystyle \sec\left(\frac{\pi}{2} - t\right) = \csc t, wherever \sin t \ne 0.
- -\cos t. \displaystyle \frac{3\pi}{2} - t = \pi + \left(\frac{\pi}{2} - t\right), and the half turn changes the sign of sine: \displaystyle -\sin\left(\frac{\pi}{2} - t\right) = -\cos t.
- \sin t. Subtract 2\pi: \displaystyle \cos\left(t + \frac{3\pi}{2}\right) = \cos\left(t - \frac{\pi}{2}\right), which is \sin t as in Exercise 9.
- 1. It is \sin t\sin t + \cos t\cos t = \sin^2 t + \cos^2 t, and (\cos t, \sin t) is on x^2 + y^2 = 1.
- \cot t, for every t that is not an integer multiple of \displaystyle \frac{\pi}{2}. The quotient is \displaystyle \frac{\csc t}{\sec t} = \frac{\cos t}{\sin t}; the original needs \sin t \ne 0 for the top and \cos t \ne 0 for the bottom, so at \displaystyle t = \frac{\pi}{2} it is undefined even though \displaystyle \cot\frac{\pi}{2} = 0.
- t \approx 0.871 and t \approx 5.412. \displaystyle \sin 0.7 = \cos\left(\frac{\pi}{2} - 0.7\right), so \displaystyle t = \frac{\pi}{2} - 0.7; the vertical line through that point meets the circle again at its mirror across the x-axis, \displaystyle 2\pi - \left(\frac{\pi}{2} - 0.7\right) = \frac{3\pi}{2} + 0.7.
- \displaystyle t = \frac{3\pi}{10} and \displaystyle t = \frac{13\pi}{10}. \displaystyle \cot\frac{\pi}{5} = \tan\left(\frac{\pi}{2} - \frac{\pi}{5}\right) = \tan\frac{3\pi}{10}, and tangent repeats every \pi.
- \theta = 250^\circ and \theta = 290^\circ. \cos 200^\circ = \sin(90^\circ - 200^\circ) = \sin(-110^\circ); -110^\circ is coterminal with 250^\circ, and the other angle with that sine is 180^\circ - (-110^\circ) = 290^\circ.
- The terminal point of -0.5 is reached from (1, 0) by an arc of length 0.5 run clockwise. The reflection sends (1, 0) to (0, 1) and reverses direction, so the image is reached from (0, 1) by an arc of length 0.5 run counterclockwise. The point (0, 1) is at \displaystyle \frac{\pi}{2}, so the image is at \displaystyle \frac{\pi}{2} + 0.5 = \frac{\pi}{2} - (-0.5).
- Only \displaystyle 0 < t < \frac{\pi}{2}: both t and \displaystyle \frac{\pi}{2} - t must be acute angles of the triangle. The unit circle argument works for every real t, including negative ones and ones past a quarter turn, because the reflection carries any arc, of any length and either direction.
- \displaystyle \csc\left(\frac{\pi}{2} - t\right) = \frac{1}{\sin\left(\frac{\pi}{2} - t\right)} = \frac{1}{\cos t} = \sec t. Both sides are defined exactly when \cos t \ne 0, that is for \displaystyle t \ne \frac{\pi}{2} + k\pi, k any integer.
- Let \displaystyle u = \frac{\pi}{2} - t, so \displaystyle t = \frac{\pi}{2} - u. The sine identity holds for every real number, in particular for u: \displaystyle \sin\left(\frac{\pi}{2} - u\right) = \cos u. Substituting back, \displaystyle \sin t = \cos\left(\frac{\pi}{2} - t\right).
- Quadrant IV. In Quadrant II the point (x, y) has x < 0 < y; swapping gives a first coordinate that is positive and a second that is negative. Check: \displaystyle \sin\left(\frac{\pi}{2} - t\right) = \cos t < 0, and \displaystyle \cos\left(\frac{\pi}{2} - t\right) = \sin t > 0, the signs of Quadrant IV.