Advanced Trig Identity Proofs

In Grade 11 you proved some first identities using $\sin^2 x + \cos^2 x = 1$ and the quotient and reciprocal identities. Now your toolkit is much bigger: you also have the compound angle and double angle formulas. Proving identities is like a puzzle: you know where you start and where you need to end up, and you choose the moves that get you there. All angles are in radians.

Key ideas

What an identity is

An identity is an equation that is true for every value of the variable for which both sides are defined. For example, $\tan x = \dfrac{\sin x}{\cos x}$ is true for every $x$ except where $\cos x = 0$ (where neither side is defined).

An ordinary equation, like $\cos 2x = 2\cos x$, is true only for some values of $x$.

• To show an equation is not an identity, one counterexample is enough: a single value of $x$ where the two sides are different.
• To show an equation is an identity, checking values or graphs isn’t enough, because you can’t check every $x$. You need a proof.

Your toolkit

Type	Identities
Pythagorean	$\sin^2 x + \cos^2 x = 1$, $\quad 1 + \tan^2 x = \sec^2 x$, $\quad 1 + \cot^2 x = \csc^2 x$
Quotient	$\tan x = \dfrac{\sin x}{\cos x}$, $\quad \cot x = \dfrac{\cos x}{\sin x}$
Reciprocal	$\csc x = \dfrac{1}{\sin x}$, $\quad \sec x = \dfrac{1}{\cos x}$, $\quad \cot x = \dfrac{1}{\tan x}$
Compound angle	$\sin(a \pm b) = \sin a\cos b \pm \cos a\sin b$, $\quad \cos(a \pm b) = \cos a\cos b \mp \sin a\sin b$
Double angle	$\sin 2x = 2\sin x\cos x$, $\quad \cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$

The second and third Pythagorean identities come from the first. Divide every term of $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$ to get $\tan^2 x + 1 = \sec^2 x$, or by $\sin^2 x$ to get $1 + \cot^2 x = \csc^2 x$.

How to lay out a proof

In Ontario classes, a proof is usually laid out in two columns: the left side (L.S.) and the right side (R.S.). You work down each column separately until both columns reach the same expression, then write “L.S. = R.S.”

$$\def\arraystretch{1.6} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \text{(the left side)} & \text{(the right side)} \\ = \ldots & = \ldots \\ = \text{same expression} & = \text{same expression} \end{array}$$ $$\therefore\ \text{L.S.} = \text{R.S.}$$

The golden rule: never move anything across the equals sign. You’re trying to show the two sides are equal, so you can’t start by assuming it. That means no adding to both sides, no cross-multiplying, and no squaring both sides. Often you only need to work on one side; that’s fine too.

Strategies

1. Start with the more complicated side. It usually has more to simplify.
2. Rewrite everything in terms of sine and cosine.
3. Expand any compound or double angles ($\sin 2x$, $\cos(x + y)$) so everything is in terms of single angles.
4. Combine fractions using a common denominator.
5. Factor: common factors, difference of squares, trinomials.
6. Look for $\sin^2 x + \cos^2 x$, or a rearranged form like $1 - \cos^2 x = \sin^2 x$, to substitute.
7. Keep your eye on the goal: if the other side has only cosines, use $\cos 2x = 2\cos^2 x - 1$ rather than another form.
8. If you get stuck, try working on the other side.

Checking with technology

Before (or after) a proof, graph both sides on graphing technology in radian mode. If the graphs are identical, you know you’re trying to prove something true. If they’re different, look for a value of $x$ where they disagree, and you have your counterexample.

Worked examples

Example 1: A counterexample

Is $\cos 2x = 2\cos x$ an identity?

Solution. Try $x = 0$:

$$\text{L.S.} = \cos 0 = 1 \qquad\qquad \text{R.S.} = 2\cos 0 = 2$$

Since $1 \ne 2$, the equation is false for $x = 0$, so it is not an identity.

The graphs of $y = \cos 2x$ and $y = 2\cos x$ are different, so the equation is not an identity.

Notice that the two graphs do cross at two points. At those values of $x$ the equation happens to be true, but an identity has to be true everywhere.

Example 2: Working on one side

Prove that $\sec x - \cos x = \sin x\tan x$.

Solution. The left side has a reciprocal function and a subtraction, so start there. The right side stays as it is.

$$\def\arraystretch{1.8} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \sec x - \cos x & \sin x\tan x \\ = \dfrac{1}{\cos x} - \cos x & \\ = \dfrac{1}{\cos x} - \dfrac{\cos^2 x}{\cos x} & \\ = \dfrac{1 - \cos^2 x}{\cos x} & \\ = \dfrac{\sin^2 x}{\cos x} & \\ = \sin x \cdot \dfrac{\sin x}{\cos x} & \\ = \sin x\tan x & \end{array}$$ $$\therefore\ \text{L.S.} = \text{R.S.}$$

Example 3: Working on both sides

Prove that $\tan x + \cot x = 2\csc 2x$.

Solution. Both sides can be rewritten in sine and cosine. Work on each column until they meet.

$$\def\arraystretch{1.8} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \tan x + \cot x & 2\csc 2x \\ = \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} & = \dfrac{2}{\sin 2x} \\ = \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} & = \dfrac{2}{2\sin x\cos x} \\ = \dfrac{1}{\sin x\cos x} & = \dfrac{1}{\sin x\cos x} \end{array}$$ $$\therefore\ \text{L.S.} = \text{R.S.}$$

Example 4: Using compound angles

Prove that $\sin(x + y)\sin(x - y) = \sin^2 x - \sin^2 y$.

Solution. Expand the left side. The product has the form $(A + B)(A - B) = A^2 - B^2$.

$$\def\arraystretch{1.6} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \sin(x + y)\sin(x - y) & \sin^2 x - \sin^2 y \\ = (\sin x\cos y + \cos x\sin y)(\sin x\cos y - \cos x\sin y) & \\ = \sin^2 x\cos^2 y - \cos^2 x\sin^2 y & \\ = \sin^2 x(1 - \sin^2 y) - (1 - \sin^2 x)\sin^2 y & \\ = \sin^2 x - \sin^2 x\sin^2 y - \sin^2 y + \sin^2 x\sin^2 y & \\ = \sin^2 x - \sin^2 y & \end{array}$$ $$\therefore\ \text{L.S.} = \text{R.S.}$$

The right side only has sines, so the key move was replacing each $\cos^2$ with $1 - \sin^2$.

Common mistakes

Treating the identity like an equation to solve. Moving terms across, cross-multiplying, or squaring both sides assumes the two sides are already equal, which is what you’re trying to prove. Work on each side separately.

Using a numerical check or a graph as a proof. A value or a graph that agrees is a good sign, but it isn’t a proof: there are infinitely many values you didn’t check. (A value that disagrees is a valid disproof, though.)

Distributing a trig function. $\sin(x + y) \ne \sin x + \sin y$ and $\sin 2x \ne 2\sin x$. Use the compound and double angle formulas.

Cancelling terms instead of factors. In $\dfrac{\sin^2 x + \sin x}{\sin x}$, you can’t cancel the $\sin x$ from just one term. Factor first: $\dfrac{\sin x(\sin x + 1)}{\sin x} = \sin x + 1$.

Forgetting where the identity is defined. An identity only has to hold where both sides are defined. In Example 3, neither side exists when $\sin 2x = 0$ (for example at $x = 0$ or $x = \dfrac{\pi}{2}$), and that’s fine.

Stopping too early. Keep going until both columns show exactly the same expression, then write the conclusion “L.S. = R.S.”

Practice

1. (Warm-up) Use a counterexample to show that $(\sin x + \cos x)^2 = 1$ is not an identity.

Solution

Try $x = \dfrac{\pi}{4}$: $\text{L.S.} = \left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}\right)^2 = (\sqrt{2})^2 = 2$, but $\text{R.S.} = 1$. Since $2 \ne 1$, it isn’t an identity.

(In fact, $(\sin x + \cos x)^2 = 1 + \sin 2x$, which is only equal to $1$ when $\sin 2x = 0$.)

2. (Warm-up) Prove that $\cot x\sec x = \csc x$.

Solution

$$\def\arraystretch{1.8} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \cot x\sec x & \csc x \\ = \dfrac{\cos x}{\sin x}\cdot\dfrac{1}{\cos x} & = \dfrac{1}{\sin x} \\ = \dfrac{1}{\sin x} & \end{array}$$

$\therefore$ L.S. = R.S.

3. (Core) Prove that $\dfrac{\sin 2x}{\sin x} - \dfrac{\cos 2x}{\cos x} = \sec x$.

Solution

$$\def\arraystretch{1.8} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \dfrac{\sin 2x}{\sin x} - \dfrac{\cos 2x}{\cos x} & \sec x \\ = \dfrac{2\sin x\cos x}{\sin x} - \dfrac{2\cos^2 x - 1}{\cos x} & = \dfrac{1}{\cos x} \\ = 2\cos x - \dfrac{2\cos^2 x - 1}{\cos x} & \\ = \dfrac{2\cos^2 x - (2\cos^2 x - 1)}{\cos x} & \\ = \dfrac{1}{\cos x} & \end{array}$$

$\therefore$ L.S. = R.S.

4. (Core) Prove that $\dfrac{\sin x}{1 - \cos x} = \dfrac{1 + \cos x}{\sin x}$.

Solution

Multiply the top and bottom of the left side by $1 + \cos x$ (the “conjugate” of the denominator), which doesn’t change its value:

$$\def\arraystretch{1.8} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \dfrac{\sin x}{1 - \cos x}\cdot\dfrac{1 + \cos x}{1 + \cos x} & \dfrac{1 + \cos x}{\sin x} \\ = \dfrac{\sin x(1 + \cos x)}{1 - \cos^2 x} & \\ = \dfrac{\sin x(1 + \cos x)}{\sin^2 x} & \\ = \dfrac{1 + \cos x}{\sin x} & \end{array}$$

$\therefore$ L.S. = R.S.

5. (Core) Prove that $\cos(x + y)\cos(x - y) = \cos^2 x - \sin^2 y$.

Solution

$$\def\arraystretch{1.6} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \cos(x + y)\cos(x - y) & \cos^2 x - \sin^2 y \\ = (\cos x\cos y - \sin x\sin y)(\cos x\cos y + \sin x\sin y) & \\ = \cos^2 x\cos^2 y - \sin^2 x\sin^2 y & \\ = \cos^2 x(1 - \sin^2 y) - (1 - \cos^2 x)\sin^2 y & \\ = \cos^2 x - \cos^2 x\sin^2 y - \sin^2 y + \cos^2 x\sin^2 y & \\ = \cos^2 x - \sin^2 y & \end{array}$$

$\therefore$ L.S. = R.S.

6. (Core) Prove that $\sin\left(\dfrac{\pi}{4} + x\right) + \sin\left(\dfrac{\pi}{4} - x\right) = \sqrt{2}\cos x$.

Solution

$$\def\arraystretch{1.8} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \sin\left(\dfrac{\pi}{4} + x\right) + \sin\left(\dfrac{\pi}{4} - x\right) & \sqrt{2}\cos x \\ = \sin\dfrac{\pi}{4}\cos x + \cos\dfrac{\pi}{4}\sin x + \sin\dfrac{\pi}{4}\cos x - \cos\dfrac{\pi}{4}\sin x & \\ = 2\sin\dfrac{\pi}{4}\cos x & \\ = 2\left(\dfrac{\sqrt{2}}{2}\right)\cos x & \\ = \sqrt{2}\cos x & \end{array}$$

$\therefore$ L.S. = R.S.

7. (Core) Prove that $\cos 2x = \dfrac{1 - \tan^2 x}{1 + \tan^2 x}$.

Solution

Start with the right side. Use $1 + \tan^2 x = \sec^2 x$, then multiply by $\cos^2 x$:

$$\def\arraystretch{1.8} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \cos 2x & \dfrac{1 - \tan^2 x}{1 + \tan^2 x} \\ & = \dfrac{1 - \tan^2 x}{\sec^2 x} \\ & = \left(1 - \dfrac{\sin^2 x}{\cos^2 x}\right)\cos^2 x \\ & = \cos^2 x - \sin^2 x \\ & = \cos 2x \end{array}$$

$\therefore$ L.S. = R.S.

8. (Challenge) Prove that $\dfrac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x$.

Solution

In the numerator, use $\cos 2x = 1 - 2\sin^2 x$ so the $1$s cancel. In the denominator, use $\cos 2x = 2\cos^2 x - 1$ for the same reason. Then factor.

$$\def\arraystretch{1.8} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \dfrac{1 - (1 - 2\sin^2 x) + 2\sin x\cos x}{1 + (2\cos^2 x - 1) + 2\sin x\cos x} & \tan x \\ = \dfrac{2\sin^2 x + 2\sin x\cos x}{2\cos^2 x + 2\sin x\cos x} & \\ = \dfrac{2\sin x(\sin x + \cos x)}{2\cos x(\cos x + \sin x)} & \\ = \dfrac{\sin x}{\cos x} & \\ = \tan x & \end{array}$$

$\therefore$ L.S. = R.S.

9. (Challenge) Prove that $\csc 2x + \cot 2x = \cot x$.

Solution

$$\def\arraystretch{1.8} \begin{array}{l|l} \text{L.S.} & \text{R.S.} \\ \hline \dfrac{1}{\sin 2x} + \dfrac{\cos 2x}{\sin 2x} & \cot x \\ = \dfrac{1 + \cos 2x}{\sin 2x} & = \dfrac{\cos x}{\sin x} \\ = \dfrac{1 + 2\cos^2 x - 1}{2\sin x\cos x} & \\ = \dfrac{2\cos^2 x}{2\sin x\cos x} & \\ = \dfrac{\cos x}{\sin x} & \end{array}$$

$\therefore$ L.S. = R.S.

Check with $x = \dfrac{\pi}{6}$: $\csc\dfrac{\pi}{3} + \cot\dfrac{\pi}{3} = \dfrac{2}{\sqrt{3}} + \dfrac{1}{\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$, and $\cot\dfrac{\pi}{6} = \sqrt{3}$. ✓