Compound Angle Formulas for sin, cos and tan | NIMCET

Quick answer: For any angles A and B: $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$, $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$ and $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$. They give exact values such as $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$, and they show that $a\sin x + b\cos x$ always lies between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$.

An angle written as a sum or difference of two angles, such as $A + B$ or $45^\circ - 30^\circ$, is called a compound angle. The formulas below are used constantly in NIMCET: for exact values, simplifying expressions, finding maxima and minima, and later in calculus. If you are unsure about quadrant signs, revise trigonometric ratios and allied angles first.

The six formulas

A way to remember the signs: sine keeps the sign (sin of a sum has a plus), while cosine flips it (cos of a sum has a minus).

Where the tan formula comes from

Divide $\sin(A + B)$ by $\cos(A + B)$, then divide the numerator and denominator by $\cos A\cos B$:

$\tan(A + B) = \frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B} = \frac{\tan A + \tan B}{1 - \tan A\tan B}$

It is valid only when $\cos A$, $\cos B$ and $\cos(A + B)$ are non-zero.

Formulas derived from these

Exact values of 15° and 75°

Write $15^\circ = 45^\circ - 30^\circ$ and $75^\circ = 45^\circ + 30^\circ$.

$\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$

$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}$

The range of a sin x + b cos x

Write $a = r\cos\alpha$ and $b = r\sin\alpha$, where $r = \sqrt{a^2 + b^2}$. Then:

$a\sin x + b\cos x = r(\sin x\cos\alpha + \cos x\sin\alpha) = r\sin(x + \alpha)$

Since $\sin(x + \alpha)$ lies between $-1$ and $1$:

$-\sqrt{a^2 + b^2} \le a\sin x + b\cos x \le \sqrt{a^2 + b^2}$

For example, $3\sin x + 4\cos x$ has maximum 5 and minimum $-5$. This result is one of the most frequently used consequences of the compound angle formulas.

Worked examples

Example 1: tan(A + B) from given ratios

If $\tan A = \frac{1}{2}$ and $\tan B = \frac{1}{3}$, with A and B acute, find $A + B$.

Step 1: Apply the formula.

$\tan(A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \frac{5/6}{5/6} = 1$

Step 2: A and B are acute, so $0 < A + B < 180^\circ$, and $\tan(A + B) = 1$ gives $A + B = 45^\circ$.

Example 2: a product that becomes a difference of squares

Evaluate $\sin^2 52.5^\circ - \sin^2 22.5^\circ$.

Step 1: Use $\sin^2 A - \sin^2 B = \sin(A + B)\sin(A - B)$.

Step 2: Here $A + B = 75^\circ$ and $A - B = 30^\circ$:

$\sin 75^\circ \sin 30^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} \times \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{8}$

Example 3: A + B = 45°

If $A + B = 45^\circ$, find $(1 + \tan A)(1 + \tan B)$.

Step 1: $\tan(A + B) = 1$, so $\frac{\tan A + \tan B}{1 - \tan A\tan B} = 1$.

Step 2: This gives $\tan A + \tan B + \tan A\tan B = 1$.

Step 3: Expand the product:

$(1 + \tan A)(1 + \tan B) = 1 + \tan A + \tan B + \tan A\tan B = 1 + 1 = 2$

Check: With $A = 0$, $B = 45^\circ$: $(1)(2) = 2$ ✓.

Example 4: the maximum of an expression

Find the maximum value of $\sin x + \sqrt{3}\cos x$.

Here $a = 1$, $b = \sqrt{3}$, so the maximum is $\sqrt{1 + 3} = 2$. It occurs when $x + 60^\circ = 90^\circ$, that is, $x = 30^\circ$.

Summary table

Exam tips

• Look for angles that combine into 30°, 45°, 60° or 90°; that is usually the intended route.
• When you see $\sin^2 A - \sin^2 B$, think "$\sin(A + B)\sin(A - B)$" straight away.
• For any question on the maximum or minimum of $a\sin x + b\cos x + c$, the answer is $c \pm \sqrt{a^2 + b^2}$.
• After using the tan formula, check the quadrant before converting $\tan(A + B)$ back into an angle.

More solved problems are in the NIMCET Trigonometry chapter.

Common mistakes

• Writing $\sin(A + B) = \sin A + \sin B$. The compound angle formulas are not linear.
• Getting the cos signs the wrong way round: $\cos(A + B)$ has a minus sign between the products.
• Forgetting that $\tan(A + B)$ is undefined when $\tan A\tan B = 1$, which means $A + B$ is an odd multiple of 90°.
• Using $\sqrt{a + b}$ instead of $\sqrt{a^2 + b^2}$ for the maximum of $a\sin x + b\cos x$.

Frequently Asked Questions

What is the formula for sin(A + B)?

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

What is the formula for cos(A − B)?

cos(A − B) = cos A cos B + sin A sin B. Note the plus sign.

How do you find sin 75° without a calculator?

Write 75° = 45° + 30° and use the sin(A + B) formula. This gives (√6 + √2)/4.

What is the maximum value of 3 sin x + 4 cos x?

It is √(3² + 4²) = 5, and the minimum is −5.

If A + B = 45°, what is (1 + tan A)(1 + tan B)?

It equals 2, because tan(A + B) = 1 leads to tan A + tan B + tan A tan B = 1.