Trig Addition and Subtraction Formulas

A standard trig table gives you values for 30°, 45°, 60°, and 90°. But what if you need the sine of 75°, or the cosine of 15°? The addition and subtraction formulas let you compute exact trig values for many angles that don't appear in the table — by expressing them as sums or differences of angles you already know.

Beyond exact values, these formulas are essential tools in calculus, physics, and engineering: they're used to derive the derivatives of trig functions, simplify wave equations, and prove other identities.

The Formulas

Here are all six. They're worth memorizing — or at least knowing how to reconstruct quickly.

Sine: $$\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)$$ $$\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)$$

Cosine: $$\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)$$ $$\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)$$

Tangent: $$\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}$$ $$\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)}$$

Two memory notes: for cosine, the sign in the formula is opposite the sign in the angle (addition uses a minus sign, subtraction uses a plus). For sine, the sign matches. For tangent, the numerator sign matches the angle, and the denominator sign is opposite.

Where They Come From

These formulas aren't arbitrary — they follow from basic geometry. Here's a clean derivation that requires only the distance formula and one trig identity.

Step 1: Derive cos(a − b)

Place two points on the unit circle:

• \(P = (\cos a,\ \sin a)\)
• \(Q = (\cos b,\ \sin b)\)

The squared distance between them is:

$$d^2 = (\cos a - \cos b)^2 + (\sin a - \sin b)^2$$

Expand:

$$= \cos^2 a - 2\cos a\cos b + \cos^2 b + \sin^2 a - 2\sin a\sin b + \sin^2 b$$

Group using the Pythagorean identity (\(\cos^2\theta + \sin^2\theta = 1\)):

$$= 1 + 1 - 2(\cos a\cos b + \sin a\sin b) = 2 - 2(\cos a\cos b + \sin a\sin b)$$

Now rotate both points by \(-b\). Rotation preserves distance, and after rotating:

• \(P\) moves to \(P' = (\cos(a-b),\ \sin(a-b))\)
• \(Q\) moves to \(Q' = (1,\ 0)\)

So the squared distance is also:

$$d^2 = (\cos(a-b) - 1)^2 + \sin^2(a-b) = 2 - 2\cos(a-b)$$

Setting the two expressions equal:

$$2 - 2\cos(a-b) = 2 - 2(\cos a\cos b + \sin a\sin b)$$

$$\boxed{\cos(a-b) = \cos a\cos b + \sin a\sin b}$$

Step 2: Derive the remaining five from cos(a − b)

cos(a + b): Replace \(b\) with \(-b\) and use \(\cos(-b) = \cos b\), \(\sin(-b) = -\sin b\):

$$\cos(a+b) = \cos a\cos(-b) + \sin a\sin(-b) = \cos a\cos b - \sin a\sin b$$

sin(a + b): Use the co-function identity \(\sin\theta = \cos!\left(\frac{\pi}{2} - \theta\right)\):

$$\sin(a+b) = \cos!\left(\frac{\pi}{2} - (a+b)\right) = \cos!\left(\left(\frac{\pi}{2}-a\right) - b\right)$$

Apply the \(\cos(\alpha - \beta)\) formula with \(\alpha = \frac{\pi}{2}-a\):

$$= \cos!\left(\frac{\pi}{2}-a\right)\cos b + \sin!\left(\frac{\pi}{2}-a\right)\sin b = \sin a\cos b + \cos a\sin b$$

sin(a − b): Replace \(b\) with \(-b\) in sin(a + b):

$$\sin(a-b) = \sin a\cos b - \cos a\sin b$$

tan(a ± b): Divide the sine formula by the cosine formula:

$$\tan(a+b) = \frac{\sin(a+b)}{\cos(a+b)} = \frac{\sin a\cos b + \cos a\sin b}{\cos a\cos b - \sin a\sin b}$$

Divide numerator and denominator by \(\cos a\cos b\):

$$= \frac{\tan a + \tan b}{1 - \tan a\tan b}$$

The subtraction version follows by the same step with \(-b\).

Finding Exact Trig Values

This is one of the most useful applications. If you can write an angle as a sum or difference of 30°, 45°, or 60°, you can find its exact trig value.

Example 1: Find sin(75°)

Write \(75° = 45° + 30°\):

$$\sin(75°) = \sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30°$$

$$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$

Example 2: Find cos(15°)

Write \(15° = 45° - 30°\):

$$\cos(15°) = \cos(45° - 30°) = \cos 45°\cos 30° + \sin 45°\sin 30°$$

$$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6}+\sqrt{2}}{4}$$

Notice: \(\sin(75°) = \cos(15°)\). That's expected — 75° and 15° are complementary, and \(\sin\theta = \cos(90°-\theta)\).

Example 3: Find sin(15°)

$$\sin(15°) = \sin(45° - 30°) = \sin 45°\cos 30° - \cos 45°\sin 30°$$

$$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6}-\sqrt{2}}{4}$$

Example 4: Find cos(105°)

Write \(105° = 60° + 45°\):

$$\cos(105°) = \cos(60° + 45°) = \cos 60°\cos 45° - \sin 60°\sin 45°$$

$$= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2}-\sqrt{6}}{4}$$

The negative result makes sense — 105° is in the second quadrant, where cosine is negative.

Double Angle Formulas

A special case of the addition formulas: set \(b = a\). This gives you the double angle formulas, which appear constantly in calculus and physics.

Sine double angle:

$$\sin(2a) = \sin(a+a) = \sin a\cos a + \cos a\sin a = 2\sin a\cos a$$

Cosine double angle:

$$\cos(2a) = \cos(a+a) = \cos^2 a - \sin^2 a$$

This can also be rewritten using the Pythagorean identity \(\sin^2 a + \cos^2 a = 1\):

$$\cos(2a) = 2\cos^2 a - 1 \qquad \text{or} \qquad \cos(2a) = 1 - 2\sin^2 a$$

All three forms of the cosine double angle formula are useful — which one you use depends on what you're trying to simplify.

Tangent double angle:

$$\tan(2a) = \frac{2\tan a}{1 - \tan^2 a}$$

More Worked Examples

Example 5: Simplify an expression

Simplify \(\sin(x+\pi)\).

$$\sin(x + \pi) = \sin x\cos\pi + \cos x\sin\pi = \sin x(-1) + \cos x(0) = -\sin x$$

So \(\sin(x+\pi) = -\sin x\) — shifting the sine function by \(\pi\) flips it.

Example 6: Prove an identity

Show that \(\cos(x - \pi/2) = \sin x\).

$$\cos!\left(x - \frac{\pi}{2}\right) = \cos x\cos\frac{\pi}{2} + \sin x\sin\frac{\pi}{2} = \cos x(0) + \sin x(1) = \sin x \quad \checkmark$$

Example 7: Find sin(2θ) given other information

If \(\sin\theta = \frac{3}{5}\) and \(\theta\) is in the first quadrant, find \(\sin(2\theta)\).

From \(\sin\theta = \frac{3}{5}\), use the Pythagorean identity to find \(\cos\theta = \frac{4}{5}\) (positive in quadrant I).

$$\sin(2\theta) = 2\sin\theta\cos\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}$$

Practice Problems

Find the exact value of \(\sin(105°)\). Write \(105° = 60° + 45°\).

Show answer\(\sin(60°+45°) = \sin 60°\cos 45° + \cos 60°\sin 45° = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}\)

Find the exact value of \(\cos(75°)\). Write \(75° = 45° + 30°\).

Show answer\(\cos(45°+30°) = \cos 45°\cos 30° - \sin 45°\sin 30° = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}\)

Simplify \(\cos(x + \pi)\).

Show answer\(\cos(x+\pi) = \cos x\cos\pi - \sin x\sin\pi = \cos x(-1) - \sin x(0) = -\cos x\)

If \(\cos\theta = \frac{5}{13}\) and \(\theta\) is in quadrant I, find \(\cos(2\theta)\).

Show answer\(\sin\theta = \frac{12}{13}\) (from Pythagorean theorem). \(\cos(2\theta) = \cos^2\theta - \sin^2\theta = \frac{25}{169} - \frac{144}{169} = -\frac{119}{169}\)

Prove that \(\sin(a+b) + \sin(a-b) = 2\sin a\cos b\).

Show answerExpand both: \((\sin a\cos b + \cos a\sin b) + (\sin a\cos b - \cos a\sin b) = 2\sin a\cos b\). The \(\cos a\sin b\) terms cancel. ✓

Find \(\tan(75°)\) using the addition formula. Write \(75° = 45° + 30°\).

Show answer\(\tan(45°+30°) = \frac{\tan 45° + \tan 30°}{1 - \tan 45°\tan 30°} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}\). Rationalize: \(\frac{(\sqrt{3}+1)^2}{(\sqrt{3})^2-1^2} = \frac{3+2\sqrt{3}+1}{2} = 2+\sqrt{3}\).