What an identity is
The equals sign in 2\cos t = 1 asks which angles work. The equals sign in \cos(-t) = \cos t claims something about every angle at once. How do you tell the two apart, and what does “every angle” mean when one side has no value at some angles?
Where you are 13 in Trigonometric identities
Why this exists
The unit circle lessons used one equals sign for two different kinds of statement. One kind asks a question: 2\cos t = 1 holds at \displaystyle t = \frac{\pi}{3}, at \displaystyle t = -\frac{\pi}{3} and at every angle that differs from those by full turns in either direction, and fails everywhere else, so the work is to find those angles. The other kind states a fact: \cos(-t) = \cos t and \sin(t + 2\pi) = \sin t hold for every angle, so there is nothing to solve. Instead, either side may be swapped for the other wherever it appears.
This unit is built on the second kind. Simplifying, verifying and solving all work by replacing one side of a known identity with the other, which is safe only if the statement really holds for every input.
The idea is not new. A linear equation whose variable cancels into a true statement holds for every number, and polynomial identities are proved by expanding. Trigonometry adds one feature: tangent, secant, cosecant and cotangent have no value at some angles, so the two sides of a statement need not exist at the same inputs.
Definition
An equation in a variable claims that two expressions have the same value. Only a value at which both sides are defined can be tested, and the set of those values is the equation’s common domain.
- An equation is an identity if it is true for every value in its common domain.
- An equation is a conditional equation if it is true for some values in its common domain and false for others. The values where it is true are its solutions.
- An equation true for no value at all has no solution.
Some books write \equiv in place of = to flag an identity, as in \cos(-t) \equiv \cos t. This lesson uses = and says in words when a statement is an identity.
“Where both sides are defined” is part of the definition, not a footnote. An input at which one side has no value is never tested, so it can never be a counterexample. When an identity is said to hold “on” a set, that set is its common domain.
Intuition
An equation f(x) = g(x) asks where the graphs of y = f(x) and y = g(x) meet. For a conditional equation the graphs cross at separate points and part in between. In Figure 1 the sine wave crosses the line \displaystyle y = \frac{1}{2} twice in every turn, and those crossings are the solutions of \displaystyle \sin x = \frac{1}{2}.
For an identity the two graphs are the same graph: at every input where both sides exist they give the same point, so the curves lie exactly on top of each other. If one side is undefined at some inputs, its graph has holes there, and the picture is the other graph with those points punched out. In Figure 2, y = \tan x \cos x is the sine wave except at the inputs where \cos x = 0, where \tan x has no value and the graph has no point.
Two curves apart at some input give a counterexample; two curves together are not a proof, since a graph covers one window at the width of a pen line. In Figure 3, y = \sin 2x and y = 2\sin x meet at every multiple of \pi, so testing only 0, \pi and 2\pi finds agreement every time, yet the curves are far apart in between.
Results
What an identity claims. f(x) = g(x) is an identity when f(a) = g(a) for every a at which both sides are defined. In pictures: the graphs of y = f(x) and y = g(x) have the same point at every such input.
One counterexample is enough to disprove. If a single value a in the common domain gives f(a) \neq g(a), the equation is not an identity.
No number of checks proves. Whatever values are checked, some equation that is not an identity passes every one of those checks. Proving an identity takes an argument that covers every input at once.
The first trigonometric identities. For every real number t,
and for every t with \cos t \neq 0, that is, \displaystyle t \neq \frac{\pi}{2} + k\pi for every integer k,
Why it is true
The trigonometric identities. The angle t lands at a point (x, y) on the unit circle, with \cos t = x and \sin t = y. The angle -t is the same turn in the other direction, so it lands at the mirror image (x, -y) across the x-axis. Reading off coordinates, \cos(-t) = x = \cos t and \sin(-t) = -y = -\sin t. Nothing here depends on which t was chosen, and sine and cosine are defined at every real number, so both hold for every real t. The angle t + 2\pi is t followed by one full turn, which ends at the same point, so sine and cosine are unchanged. Tangent was defined as \displaystyle \frac{y}{x}, so wherever x \neq 0, \displaystyle \tan t = \frac{y}{x} = \frac{\sin t}{\cos t}. That one is true by definition; the next lesson, Reciprocal and quotient identities, names it and its partner for cotangent.
One counterexample. An identity claims that something happens at every input in its common domain. A claim about every input is false as soon as one input fails.
Why checks cannot prove. Suppose f(x) = g(x) is an identity and someone checks it at the values a_1, a_2, \dots, a_n. Build
The product is 0 at each a_i, so h(a_i) = f(a_i) = g(a_i), and the equation h(x) = g(x) passes every one of the same checks. At any other input no factor is zero, so h(x) \neq f(x) = g(x) there, and h(x) = g(x) is not an identity. The checks cannot tell the two equations apart.
Worked examples
Identity or conditional
Classify each equation. For a conditional equation, give a value where it holds and one where it fails.
(a) \sin(t + 2\pi) = \sin t (b) 2\cos t = 1 (c) \cos(t + \pi) = \cos t
Solution(a) Both sides are defined at every real t, and adding a full turn returns to the same point on the circle, so the sines match for every t. Identity.
(b) The equation says \displaystyle \cos t = \frac{1}{2}. At \displaystyle t = \frac{\pi}{3} the left side is \displaystyle 2 \cdot \frac{1}{2} = 1, true. At t = 0 it is 2 \cdot 1 = 2 \neq 1, false. Conditional.
(c) The angle t + \pi is a half turn past t, so it lands at the opposite point (-x, -y) and \cos(t + \pi) = -\cos t. The equation becomes -\cos t = \cos t, which means \cos t = 0. At \displaystyle t = \frac{\pi}{2} both sides are 0, true. At t = 0 the left side is \cos \pi = -1 and the right side is 1, false. Conditional, with solutions \displaystyle t = \frac{\pi}{2} + k\pi for every integer k.
One counterexample
Natalia suspects that \sin 2t = 2\sin t is an identity. She checked t = 0, t = \pi and t = 2\pi, and both sides matched each time. Settle the question.
SolutionAt her three values both sides are 0, since \sin 0 = \sin 2\pi = \sin 4\pi = 0 and 2\sin 0 = 2\sin \pi = 2\sin 2\pi = 0. Every check landed where the graphs in Figure 3 cross.
Try \displaystyle t = \frac{\pi}{2}, where both sides are defined:
Since 0 \neq 2, the equation is not an identity. One value settles it, and her three matches count for nothing. Even without a calculation: 2\sin t reaches 2, while \sin 2t never exceeds 1.
A cancelled factor
Is \displaystyle \frac{x^2 - 5x}{x} = x - 5 an identity? On what set?
SolutionThe left side is undefined at x = 0, where its denominator is zero; the right side is defined everywhere. The common domain is every real x \neq 0. For such x,
and the cancellation is allowed because x \neq 0. So the equation is an identity on x \neq 0.
At x = 0 the right side is -5 and the left side has no value. That is not a counterexample, because 0 is outside the common domain. The graph of the left side is the line y = x - 5 with one point, (0, -5), removed.
A product with gaps
Show that \tan t \cos t = \sin t is an identity, and find its common domain.
SolutionThe left side needs \tan t, which is defined where \cos t \neq 0, that is, at \displaystyle t \neq \frac{\pi}{2} + k\pi for every integer k. The right side is defined everywhere. So the common domain is \displaystyle t \neq \frac{\pi}{2} + k\pi.
For any such t, \displaystyle \tan t = \frac{\sin t}{\cos t}, so
and the cosines cancel because \cos t \neq 0. This holds at every input of the common domain, so it is an identity. Figure 2 is its picture: the sine wave with a hole at each excluded input.
Two sides with different gaps
Is \displaystyle \frac{\sin t}{\tan t} = \cos t an identity? On what set?
SolutionThe left side needs \tan t to be defined, so \cos t \neq 0, and needs it to be nonzero. Since \displaystyle \tan t = \frac{\sin t}{\cos t}, it is zero exactly where \sin t = 0. Cosine is zero at \displaystyle \frac{\pi}{2} + k\pi and sine at k\pi; together those are the multiples of \displaystyle \frac{\pi}{2}. The right side is defined everywhere, so the common domain is \displaystyle t \neq \frac{k\pi}{2} for every integer k. For such t,
with \sin t \neq 0 allowing the cancellation. It is an identity on \displaystyle t \neq \frac{k\pi}{2}.
Check. At \displaystyle t = \frac{\pi}{3}: \displaystyle \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} and \displaystyle \tan\frac{\pi}{3} = \sqrt{3}, so the left side is \displaystyle \frac{1}{2} = \cos\frac{\pi}{3}. At t = 0 the left side is \displaystyle \frac{0}{0} and the right side is 1; at \displaystyle t = \frac{\pi}{2} the left side has no value and the right side is 0. Neither is a counterexample, since neither input is in the common domain.
A graph that fits too well
Tobenna graphs y = \sin t and \displaystyle y = t - \frac{t^3}{6} for -1 \le t \le 1 and sees a single curve. Is \displaystyle \sin t = t - \frac{t^3}{6} an identity?
SolutionAt t = 1 the sides are \sin 1 \approx 0.8415 and \displaystyle 1 - \frac{1}{6} \approx 0.8333. They differ by less than 0.01, thinner than a pen line on a small graph, so the window cannot show the gap.
Look outside the window. At t = \pi,
Both sides are defined there and differ, so it is not an identity. The right side is a cubic that grows without bound, while \sin t stays between -1 and 1.
Using identities to solve an equation
Solve \sin(t + 2\pi) = \cos(-t) for 0 \le t < 2\pi.
SolutionBoth \sin(t + 2\pi) = \sin t and \cos(-t) = \cos t hold for every real t, so replacing each side by its partner keeps exactly the same solutions:
This is conditional, not an identity: at t = 0 it reads 0 = 1. Sine and cosine are never both zero, because (0, 0) is not on the unit circle, so any solution has \cos t \neq 0 and dividing by \cos t is allowed: \tan t = 1. In 0 \le t < 2\pi that gives
Check. At \displaystyle \frac{\pi}{4} both sine and cosine are \displaystyle \frac{\sqrt{2}}{2}; at \displaystyle \frac{5\pi}{4} both are \displaystyle -\frac{\sqrt{2}}{2}.
Boundaries
Only the common domain counts. An identity says nothing at an input where either side is undefined. It follows that two sides of an identity can still be different functions: \tan t \cos t and \sin t agree wherever both exist, but the first has gaps. Replacing one by the other inside a bigger expression can add inputs to the domain, so when solving, a value excluded at the start stays excluded (Exercise 22).
A stated set changes the claim. If a problem restricts the variable, “identity” means true at every value in that set. \displaystyle \sqrt{x^2} = x holds for every x \ge 0, but on all real numbers it is not an identity, since x = -1 gives 1 \neq -1.
Checking can prove when the kind of function is known. If both sides are known in advance to be polynomials of degree at most n, then agreement at n + 1 different numbers does prove they are the same polynomial: their difference has degree at most n and n + 1 zeros, which only the zero polynomial has. Sine has no such count: \sin t is zero at every multiple of \pi, infinitely many inputs, without being the zero function.
Where this goes next
The rest of this unit collects identities and puts them to work. Reciprocal and quotient identities names \displaystyle \tan t = \frac{\sin t}{\cos t} and the matching facts for the other four functions. The Pythagorean identity turns the circle equation x^2 + y^2 = 1 into a statement about sine and cosine, and the Cofunction identities relate an angle to its complement.
Verifying an identity gives the “every input at once” argument a proof needs. With identities in hand, messy expressions shrink, and trigonometric equations are solved by rewriting them first, as in Example 7. Example 2 showed what \sin 2t is not; the double angle formula says what it is.
Exercises
Identity or conditional
Classify each equation as an identity, a conditional equation, or an equation with no solution. For a conditional equation, give one value where it holds and one where it fails.
- \sin(-t) = -\sin t
- \cos(t + 2\pi) = \cos t
- \displaystyle 2\sin t = \sqrt{3}
- \tan(-t) = \tan t
- \sin(t - \pi) = \sin t
- \cos t = 2
Counterexamples
Show that each equation is not an identity by giving one counterexample.
- \displaystyle \sin\frac{t}{2} = \frac{1}{2}\sin t
- \cos 2t = 2\cos t
- \sin(a + b) = \sin a + \sin b
- (\sin t)^2 = \sin t
- Tobenna checks \tan t = \sin t at t = 0, t = \pi and t = 2\pi, and it holds every time. Is it an identity?
Common domains
Each equation is an identity. State its common domain and show that it holds there.
- \sec t \cos t = 1
- \displaystyle \frac{x^2 - 49}{x + 7} = x - 7
- \cot t \sin t = \cos t
- \displaystyle \frac{\tan t}{\sin t} = \sec t
- \displaystyle \frac{t^2 + 1}{t^2 + 1} = 1
Graphs
- On 0 \le t < 2\pi, the graphs of y = \sin t and y = -\cos t cross twice. Is \sin t = -\cos t an identity? Find where it is true in that interval.
- List every hole in the graph of y = \tan t \cos t for -2\pi \le t \le 2\pi, as points (t, y).
- Nomvula graphs y = \cos t and \displaystyle y = 1 - \frac{t^2}{2} for -0.5 \le t \le 0.5 and sees one curve. Give a counterexample to \displaystyle \cos t = 1 - \frac{t^2}{2}.
Identities as tools
Solve each equation for 0 \le t < 2\pi.
- \displaystyle \sin(t + 2\pi) = \frac{1}{2}
- \displaystyle \tan t \cos t = \frac{\sqrt{3}}{2}
- \tan t \cos t = 1
Reasoning
- Explain why one value where the two sides differ shows that an equation is not an identity, while a thousand values where they agree do not show that it is.
- Natalia says \tan t \cos t = \sin t is not an identity, because at \displaystyle t = \frac{\pi}{2} the right side is 1 and the left side is not. Is she right?
- Using only the identity \cos(t + 2\pi) = \cos t, explain why \cos(t + 4\pi) = \cos t is an identity.
Answers
- Identity. The angle -t lands at (x, -y), so \sin(-t) = -y = -\sin t for every real t.
- Identity. A full turn returns to the same point, for every real t.
- Conditional. It holds at \displaystyle t = \frac{\pi}{3}, since \displaystyle 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}, and fails at t = 0, since \displaystyle 0 \neq \sqrt{3}.
- Conditional. \displaystyle \tan(-t) = \frac{-\sin t}{\cos t} = -\tan t, so the equation means \tan t = 0. It holds at t = 0 and fails at \displaystyle t = \frac{\pi}{4}, where the sides are -1 and 1.
- Conditional. A half turn back gives \sin(t - \pi) = -\sin t, so the equation means \sin t = 0. It holds at t = 0 and fails at \displaystyle t = \frac{\pi}{2}, where the sides are -1 and 1.
- No solution. \cos t is a coordinate of a point on the unit circle, so -1 \le \cos t \le 1 and it is never 2.
- At t = \pi: \displaystyle \sin\frac{\pi}{2} = 1, but \displaystyle \frac{1}{2}\sin\pi = 0.
- At t = 0: \cos 0 = 1, but 2\cos 0 = 2.
- At \displaystyle a = b = \frac{\pi}{2}: \sin\pi = 0, but \displaystyle \sin\frac{\pi}{2} + \sin\frac{\pi}{2} = 2.
- At \displaystyle t = -\frac{\pi}{2}: (\sin t)^2 = (-1)^2 = 1, but \sin t = -1.
- No. His values all have \sin t = 0, where both sides are 0. At \displaystyle t = \frac{\pi}{4}, \tan t = 1 but \displaystyle \sin t = \frac{\sqrt{2}}{2}.
- Common domain \displaystyle t \neq \frac{\pi}{2} + k\pi, k any integer, where \sec t is defined. There \displaystyle \sec t = \frac{1}{\cos t}, so \displaystyle \sec t \cos t = \frac{\cos t}{\cos t} = 1.
- Common domain x \neq -7. There \displaystyle \frac{(x - 7)(x + 7)}{x + 7} = x - 7, cancelling the nonzero factor x + 7.
- Common domain t \neq k\pi, k any integer, where \sin t \neq 0. There \displaystyle \cot t = \frac{\cos t}{\sin t}, so \cot t \sin t = \cos t.
- Common domain \displaystyle t \neq \frac{k\pi}{2}, k any integer: \tan t needs \cos t \neq 0 and the denominator needs \sin t \neq 0. There the left side is \displaystyle \frac{\sin t}{\cos t} \cdot \frac{1}{\sin t}, which is \displaystyle \frac{1}{\cos t} = \sec t.
- Common domain all real numbers, because t^2 + 1 \ge 1 is never zero. Any nonzero number divided by itself is 1.
- No: at t = 0 it reads 0 = -1. Where it holds, \cos t \neq 0 (else both would be zero), so \tan t = -1, giving \displaystyle t = \frac{3\pi}{4} and \displaystyle t = \frac{7\pi}{4}.
- The holes are where \cos t = 0, at height \sin t: \displaystyle \left(-\frac{3\pi}{2}, 1\right), \displaystyle \left(-\frac{\pi}{2}, -1\right), \displaystyle \left(\frac{\pi}{2}, 1\right) and \displaystyle \left(\frac{3\pi}{2}, -1\right).
- At t = \pi: \cos\pi = -1, but \displaystyle 1 - \frac{\pi^2}{2} \approx 1 - 4.9348 = -3.9348.
- The identity turns it into \displaystyle \sin t = \frac{1}{2}: \displaystyle t = \frac{\pi}{6} or \displaystyle t = \frac{5\pi}{6}.
- On its common domain the left side is \sin t, so \displaystyle \sin t = \frac{\sqrt{3}}{2}: \displaystyle t = \frac{\pi}{3} or \displaystyle t = \frac{2\pi}{3}. Both have \displaystyle \cos t = \pm\frac{1}{2} \neq 0, so both stand.
- No solution. On the common domain it means \sin t = 1, whose only value in the interval is \displaystyle t = \frac{\pi}{2}, but there \cos t = 0 and \tan t is undefined, so it is excluded.
- An identity claims equality at every input of the common domain, so one input that fails makes the claim false. Agreement at chosen inputs says nothing about the others: adding (x - a_1)\cdots(x - a_n) to one side gives an equation that passes the same checks and is not an identity.
- No. \displaystyle t = \frac{\pi}{2} is outside the common domain, since \displaystyle \tan\frac{\pi}{2} has no value, so it is not tested. At every input where both sides exist they agree (Example 4).
- The identity holds for every real input, including t + 2\pi. Applied to it, \cos\big((t + 2\pi) + 2\pi\big) = \cos(t + 2\pi), and once more, \cos(t + 2\pi) = \cos t. So \cos(t + 4\pi) = \cos t for every real t.