Verifying Trigonometric Identities, precalculus lesson

Graph $y = \tan x + \cot x$ and $y = \sec x \csc x$ on the same screen and the two curves land exactly on top of each other. That is strong evidence that they are the same function, but a screen shows only finitely many points. Verifying an identity means proving with algebra that two expressions agree at every input where both are defined, and a handful of algebraic moves does most of the work.

Key terms

Identity versus equation

Definition An identity is an equation that is true for every value of the variable at which both sides are defined. A conditional equation is true for some of those values but not for all of them. To show that an equation is not an identity, it is enough to find one value at which both sides are defined and differ.

In plain words You solve a conditional equation by finding the values that make it true. You verify an identity by showing that every allowed value makes it true.

Example The equation $\sin^2 x + \cos^2 x = 1$ is an identity, true for every real $x$ . The equation $\sin x = \cos x$ is conditional. It holds at $x = \frac{\pi}{4}$ , where both sides equal $\frac{\sqrt{2}}{2}$ , but fails at $x = 0$ , where the sides are $0$ and $1$ .

Common domain

Definition The common domain of an equation $f(x) = g(x)$ is the set of real numbers at which both $f(x)$ and $g(x)$ are defined. A trigonometric identity claims equality only on the common domain, and every step of a verification must be valid there. An input is excluded wherever some part of either side is undefined, most often because a denominator is zero, which includes the inputs where tangent, cotangent, secant or cosecant is undefined.

In plain words First find where both sides make sense. The identity only promises equality at those inputs.

Example In $\tan x \cos x = \sin x$ the left side needs $\cos x \neq 0$ , while the right side is defined for every $x$ . So the common domain is all real $x$ except $x = \frac{\pi}{2} + k\pi$ for integers $k$ . On it,

\tan x \cos x = \frac{\sin x}{\cos x} \cdot \cos x = \sin x,

so the equation is an identity. At $x = \frac{\pi}{2}$ the right side equals $1$ and the left side does not exist, which does not contradict the identity.

Rewriting one side

Definition To verify an identity, start from one side, usually the more complicated one, and rewrite it through a chain of equalities until it becomes the other side. Each step must be justified by algebra or by a known identity and must be valid on the common domain. An equally valid method rewrites each side separately until both reach the same expression. Do not operate on both sides at once, as when solving an equation, because that starts from the very equality you are trying to prove, and a step like squaring can turn a false statement such as $1 = -1$ into a true one.

In plain words Pick one side and keep rewriting it until it looks exactly like the other side. When nothing else suggests itself, write everything in terms of sines and cosines.

Example To verify $\cos x (\sec x - \cos x) = \sin^2 x$ , note that the left side needs $\cos x \neq 0$ . Start there, distribute, and use $\cos x \sec x = 1$ and the Pythagorean identity.

\begin{aligned} &\cos x (\sec x - \cos x) \\[0.4em] &\quad = \cos x \sec x - \cos^2 x \\[0.4em] &\quad = 1 - \cos^2 x = \sin^2 x. \end{aligned}

The chain starts at the left side and ends at the right side, which is the whole proof.

Factoring

Definition Factoring in a verification uses the same algebra as factoring polynomials, with expressions like $\sin x$ and $\cos x$ in the role of the variables. Common patterns are a common factor, the difference of squares

a^2 - b^2 = (a - b)(a + b),

and quadratic trinomials such as $u^2 - 3u + 2 = (u - 1)(u - 2)$ . A factor such as $\sin^2 x + \cos^2 x$ or $1 + \tan^2 x$ then simplifies by a Pythagorean identity.

In plain words Treat $\sin x$ and $\cos x$ like the letters $a$ and $b$ and factor as usual. Often one of the factors turns into $1$ or into a single function.

Example To verify $\sin^4 x - \cos^4 x = 1 - 2\cos^2 x$ for every real $x$ , read the left side as a difference of squares.

\begin{gathered} \sin^4 x - \cos^4 x \\[0.4em] = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) \\[0.4em] = \sin^2 x - \cos^2 x. \end{gathered}

Replacing $\sin^2 x$ by $1 - \cos^2 x$ gives $1 - 2\cos^2 x$ , the right side.

Common denominators

Definition A sum or difference of fractions is combined over a common denominator with

\begin{gathered} \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}, \\[0.8em] b \neq 0, \; d \neq 0. \end{gathered}

In a verification the denominators are trigonometric expressions, and the combined fraction is valid wherever none of them is zero.

In plain words Add trigonometric fractions exactly the way you add ordinary fractions. The new numerator often simplifies with a Pythagorean identity.

Example The identity from the opening, $\tan x + \cot x = \sec x \csc x$ , has common domain $x \neq \frac{k\pi}{2}$ for integers $k$ , since both $\sin x$ and $\cos x$ must be nonzero. On it,

\begin{aligned} \tan x + \cot x &= \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \\[0.8em] &= \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \\[0.8em] &= \frac{1}{\sin x \cos x} \\[0.8em] &= \sec x \csc x. \end{aligned}

Conjugates

Definition The conjugate of $a + b$ is $a - b$ , and the conjugate of $a - b$ is $a + b$ . Multiplying a fraction by $\frac{a + b}{a + b}$ , which equals $1$ wherever $a + b \neq 0$ , turns a denominator $a - b$ into $a^2 - b^2$ . With the Pythagorean identities, three such products collapse to a single term,

\begin{gathered} (1 - \sin x)(1 + \sin x) = \cos^2 x, \\[0.4em] (1 - \cos x)(1 + \cos x) = \sin^2 x, \\[0.4em] (\sec x - \tan x)(\sec x + \tan x) = 1. \end{gathered}

The last one needs $\cos x \neq 0$ .

In plain words A denominator like $1 - \sin x$ cannot be simplified as it stands. Multiply the top and the bottom by $1 + \sin x$ , and the bottom becomes $\cos^2 x$ , which often cancels with something in the numerator.

Example Verify

\frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{\cos x}.

The right side needs $\cos x \neq 0$ , and that also forces $\sin x \neq \pm 1$ . So the common domain is $x \neq \frac{\pi}{2} + k\pi$ , and on it $1 + \sin x \neq 0$ , which allows the multiplication.

\begin{aligned} \frac{\cos x}{1 - \sin x} &= \frac{\cos x (1 + \sin x)}{(1 - \sin x)(1 + \sin x)} \\[0.8em] &= \frac{\cos x (1 + \sin x)}{\cos^2 x} \\[0.8em] &= \frac{1 + \sin x}{\cos x}. \end{aligned}

Numerical evidence is not proof

Definition Checking that an equation holds at some inputs, or that two graphs appear to coincide, is evidence but not a proof that it is an identity, since an identity is a claim about infinitely many inputs. A single input in the common domain at which the two sides differ is a proof that the equation is not an identity.

In plain words Testing numbers is a good way to catch mistakes and to hunt for a counterexample. Only an algebraic argument proves an identity.

Example The equation $\sqrt{1 - \cos^2 x} = \sin x$ holds at $x = 0$ , $\frac{\pi}{6}$ , $\frac{\pi}{4}$ , $\frac{\pi}{3}$ and $\frac{\pi}{2}$ , and in fact at every $x$ in $[0, \pi]$ . Still it is not an identity. At $x = -\frac{\pi}{2}$ the left side is $\sqrt{1 - 0} = 1$ and the right side is $-1$ . The correct statement, true for every real $x$ , is

\sqrt{1 - \cos^2 x} = \lvert \sin x \rvert,

because the square root of a square is an absolute value.

Check yourself

Verify the following identity and state its common domain.

\frac{1 + \tan^2 x}{\csc^2 x} = \tan^2 x

Show the answer

The left side needs $\cos x \neq 0$ for the tangent and $\sin x \neq 0$ for the cosecant, so the common domain is $x \neq \frac{k\pi}{2}$ for integers $k$ . On it,

\begin{aligned} \frac{1 + \tan^2 x}{\csc^2 x} &= \frac{\sec^2 x}{\csc^2 x} \\[0.8em] &= \frac{1}{\cos^2 x} \cdot \sin^2 x \\[0.8em] &= \tan^2 x. \end{aligned}

Verify the identity

\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x} = 2\sec x.

Show the answer

Both sides need $\cos x \neq 0$ , and then $1 + \sin x \neq 0$ as well. Over the common denominator the left side is

\begin{aligned} &\frac{\cos^2 x + 1 + 2\sin x + \sin^2 x}{(1 + \sin x)\cos x} \\[0.8em] &\quad = \frac{2(1 + \sin x)}{(1 + \sin x)\cos x} \\[0.8em] &\quad = \frac{2}{\cos x} = 2\sec x. \end{aligned}

The equation $(\sin x + \cos x)^2 = 1$ is true at $x = 0$ , $\frac{\pi}{2}$ , $\pi$ and $\frac{3\pi}{2}$ . Is it an identity?

Show the answer

No. At $x = \frac{\pi}{4}$ the left side is $\left(\sqrt{2}\right)^2 = 2$ . Expanding shows why the checks passed, since

(\sin x + \cos x)^2 = 1 + 2\sin x \cos x,

which equals $1$ only where $\sin x \cos x = 0$ .

Verifying Trigonometric Identities

Key terms

Identity versus equation

Common domain

Rewriting one side

Factoring

Common denominators

Conjugates

Numerical evidence is not proof

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus