Solving Trigonometric Equations Using Sum and Difference Identities

The sum and difference identities let you rewrite \(\sin(A \pm B)\), \(\cos(A \pm B)\), and \(\tan(A \pm B)\) in terms of the individual angles \(A\) and \(B\). They are especially useful when an equation contains a compound angle, such as \(x + \frac{\pi}{4}\), or when the expression already looks like the right-hand side of one of these identities and can be collapsed back into a single trig function.

Before working through this lesson, it helps to be comfortable finding the exact value of trigonometric ratios for common angles like \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), and \(\frac{\pi}{3}\), since most examples rely on these values.

These are the formulas you will use throughout this topic:

\(\sin(A+B) = \sin A \cos B + \cos A \sin B\)

\(\sin(A-B) = \sin A \cos B - \cos A \sin B\)

\(\cos(A+B) = \cos A \cos B - \sin A \sin B\)

\(\cos(A-B) = \cos A \cos B + \sin A \sin B\)

\(\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}\)

\(\tan(A-B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}\)

Notice that the sine identities keep the same sign as the angle, while the cosine identities swap it. The tangent identities follow the same pattern as sine in the numerator and the opposite pattern in the denominator.

Most problems follow one of two directions:

1. Expand a compound angle. If the equation contains something like \(\sin(x + \frac{\pi}{4})\), expand it using the identity, then simplify and solve for \(x\).
2. Collapse a sum back into one identity. If the equation already looks like \(\cos x \cos B + \sin x \sin B\), recognize it as \(\cos(x - B)\), rewrite it as a single trig function, and solve.

Once the equation is reduced to a single trig ratio equal to a number, you solve it exactly the way you would solve first degree trigonometric equations, using reference angles and the appropriate interval to find every solution.

Solve \(\cos x \cos\frac{\pi}{3} + \sin x \sin\frac{\pi}{3} = \frac{1}{2}\) for \(0 \le x < 2\pi\).

Step 1: Recognize the left side matches \(\cos(A-B) = \cos A \cos B + \sin A \sin B\) with \(A = x\) and \(B = \frac{\pi}{3}\):

\(\cos\left(x - \frac{\pi}{3}\right) = \frac{1}{2}\)

Step 2: Let \(\theta = x - \frac{\pi}{3}\). Solve \(\cos\theta = \frac{1}{2}\). Cosine is positive in quadrants I and IV, so \(\theta = \frac{\pi}{3}\) or \(\theta = -\frac{\pi}{3}\) (equivalently \(\frac{5\pi}{3}\)), plus multiples of \(2\pi\).

Step 3: Substitute back \(\theta = x - \frac{\pi}{3}\) and solve for \(x\):

\(x - \frac{\pi}{3} = \frac{\pi}{3} \Rightarrow x = \frac{2\pi}{3}\)

\(x - \frac{\pi}{3} = -\frac{\pi}{3} \Rightarrow x = 0\)

Both values lie in \([0, 2\pi)\), so the solutions are \(x = 0\) and \(x = \frac{2\pi}{3}\).

Solve \(\sin\left(x + \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\) for \(0 \le x < 2\pi\).

Step 1: Let \(\theta = x + \frac{\pi}{4}\). Solve \(\sin\theta = \frac{\sqrt{2}}{2}\). Sine is positive in quadrants I and II, using the reference angle \(\frac{\pi}{4}\), so \(\theta = \frac{\pi}{4}\) or \(\theta = \frac{3\pi}{4}\), plus multiples of \(2\pi\).

Step 2: Substitute back and solve for \(x\):

\(x + \frac{\pi}{4} = \frac{\pi}{4} \Rightarrow x = 0\)

\(x + \frac{\pi}{4} = \frac{3\pi}{4} \Rightarrow x = \frac{\pi}{2}\)

Both fall inside \([0, 2\pi)\), so the solutions are \(x = 0\) and \(x = \frac{\pi}{2}\). The graph below shows \(y = \sin\left(x + \frac{\pi}{4}\right)\) crossing the value \(\frac{\sqrt{2}}{2}\) exactly at these two points.

Solve \(\tan\left(x + \frac{\pi}{4}\right) = 1\) for \(0 \le x < 2\pi\).

Step 1: Write out the identity: \(\tan\left(x + \frac{\pi}{4}\right) = \dfrac{\tan x + \tan\frac{\pi}{4}}{1 - \tan x \tan\frac{\pi}{4}} = \dfrac{\tan x + 1}{1 - \tan x}\).

Step 2: Set this equal to \(1\) and solve for \(\tan x\):

\(\dfrac{\tan x + 1}{1 - \tan x} = 1 \Rightarrow \tan x + 1 = 1 - \tan x \Rightarrow 2\tan x = 0 \Rightarrow \tan x = 0\)

Step 3: Solve \(\tan x = 0\) on \([0, 2\pi)\): \(x = 0\) and \(x = \pi\).

This matches solving \(\tan\left(x + \frac{\pi}{4}\right) = 1\) directly, since \(\tan\theta = 1\) at \(\theta = \frac{\pi}{4} + k\pi\), giving \(x = k\pi\). Whichever route you take, the equation reduces from a compound expression to something you already know how to solve as a second degree trigonometric equation whenever squaring or factoring is involved.

Watch for these pitfalls when solving with sum and difference identities:

• Mixing up the sign in the cosine identities: \(\cos(A+B)\) subtracts, while \(\cos(A-B)\) adds.
• Forgetting to shift the interval back after substituting \(\theta = x \pm B\); the bounds on \(\theta\) are not the same as the bounds on \(x\).
• Missing extra solutions by not adding \(2\pi\) (or \(\pi\) for tangent) before checking which values land inside the required interval.
• Trying to expand everything by brute force instead of first checking whether the expression already matches an identity pattern, which can save several steps.

These same equations sometimes combine with double angle expressions, so once this method feels comfortable, it's worth reviewing solving double angle trig equations to see how the two techniques work together.

Related lessons

• Trigonometry Help