Compound Angles
Digital Handwritten Lesson
New Millennium Academy · Virauta, Pokhara-17
Compound Angle Formulae
Contents
Page 1 of 5
§1 Motivating Activity
Activity: Is sin(A+B) = sinA + sinB?
We know the following standard values:
- $\sin 30° = \dfrac{1}{2}$
- $\sin 45° = \dfrac{1}{\sqrt{2}}$
- $\sin 75° = \ ?$ (to find)
Question: Can we find $\sin 75°$ simply by adding $\sin 30°$ and $\sin 45°$?
Check:
$$\sin 30° + \sin 45° = \frac{1}{2} + \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2\sqrt{2}} + \frac{2}{2\sqrt{2}} = \frac{\sqrt{2}+2}{2\sqrt{2}}$$But using a calculator:
$$\sin 75° = \frac{\sqrt{3}+1}{2\sqrt{2}}$$
⚠ Result
$\sin 75° \neq \sin 30° + \sin 45°$
📌 Observations
- $75°$ is not a standard angle, but $45°$ and $30°$ are standard angles.
- $75° = 45° + 30°$ — so $75°$ is a compound angle made from two standard angles.
- However, the sine of a sum is not equal to the sum of the sines.
📖 Conclusion
- $\sin 75° = \sin 45° + \sin 30°$ is incorrect. Although $75°$ is a compound angle of $45°$ and $30°$, the naive addition does not work.
- We need a special derivation of the compound angle formula to correctly evaluate trigonometric functions of compound angles.
Definition
Let $A$ and $B$ be any two angles. The algebraic sum or difference of $A$ and $B$, written as $A + B$ or $A - B$, is called the compound angle of $A$ and $B$.
For example, $75° = 45° + 30°$ is a compound angle formed from the two standard angles $45°$ and $30°$. Similarly, $15° = 45° - 30°$ is another compound angle.
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§2 Derivation of Compound Angle Formulae — Setup
Construction: Draw a unit circle (radius = 1 unit) with centre $O$. Define four points on its circumference:
- $S(1,0)$ — fixed point on the positive $x$-axis.
- $P(\cos A,\, \sin A)$ — a point such that $\angle POS = A$.
- $Q(\cos(A+B),\, \sin(A+B))$ — a point such that $\angle QOP = B$ and $\angle QOS = A+B$.
- $R(\cos B,\, -\sin B)$ — a point in the 4th quadrant such that $\angle SOR = B$.
§3 Chord Equality: $QS^2 = PR^2$
The central angles subtended by arc $QS$ and arc $PR$ are both equal to $A+B$:
$$\angle QOS = A+B \qquad \text{and} \qquad \angle POR = A+B$$Since equal central angles subtend equal arcs:
$$\text{Arc } QS = \text{Arc } PR$$And equal arcs in the same circle produce equal chords:
$$QS = PR \implies QS^2 = PR^2 \tag{I}$$
Page 3 of 5
§4 Derivation of $\cos(A+B)$ and $\cos(A-B)$
Step 1: Compute $QS^2$ using the distance formula.
$Q = (\cos(A+B),\; \sin(A+B))$ and $S = (1,\; 0)$, so:
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1$QS^2 = (\cos(A+B) - 1)^2 + (\sin(A+B) - 0)^2$Distance formula: $(x_2-x_1)^2+(y_2-y_1)^2$
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2$= \cos^2(A+B) - 2\cos(A+B) + 1 + \sin^2(A+B)$Expanding the squares: $(a-b)^2 = a^2 - 2ab + b^2$
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3$= \bigl[\cos^2(A+B) + \sin^2(A+B)\bigr] + 1 - 2\cos(A+B)$Rearranging terms
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4$= 1 + 1 - 2\cos(A+B)$Pythagorean identity: $\sin^2\theta + \cos^2\theta = 1$
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5$$QS^2 = 2 - 2\cos(A+B)$$
Step 2: Compute $PR^2$ using the distance formula.
$P = (\cos A,\; \sin A)$ and $R = (\cos B,\; -\sin B)$, so:
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1$PR^2 = (\cos A - \cos B)^2 + (\sin A - (-\sin B))^2$Distance formula applied to P and R
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2$= (\cos A - \cos B)^2 + (\sin A + \sin B)^2$
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3$= \cos^2 A - 2\cos A\cos B + \cos^2 B + \sin^2 A + 2\sin A\sin B + \sin^2 B$Expanding: $(a-b)^2=a^2-2ab+b^2$ and $(a+b)^2=a^2+2ab+b^2$
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4$= (\cos^2 A + \sin^2 A) + (\cos^2 B + \sin^2 B) - 2\cos A\cos B + 2\sin A\sin B$Grouping Pythagorean pairs
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5$= 1 + 1 - 2\cos A\cos B + 2\sin A\sin B$$\sin^2\theta+\cos^2\theta=1$ applied to both $A$ and $B$
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6$$PR^2 = 2 - 2\cos A\cos B + 2\sin A\sin B$$
Step 3: Equate $QS^2 = PR^2$ (from equation I).
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1$2 - 2\cos(A+B) = 2 - 2\cos A\cos B + 2\sin A\sin B$
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2$-2\cos(A+B) = -2\cos A\cos B + 2\sin A\sin B$Subtract 2 from both sides
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3$\cos(A+B) = \cos A\cos B - \sin A\sin B$Divide both sides by $-2$
Result (a)
$$\cos(A+B) = \cos A\cos B - \sin A\sin B$$
Derivation of $\cos(A-B)$:
Replace $B$ by $-B$ in the formula for $\cos(A+B)$:
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1$\cos(A-B) = \cos(A+(-B))$
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2$= \cos A\cos(-B) - \sin A\sin(-B)$Apply formula (a) with $B$ replaced by $-B$
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3$= \cos A\cdot\cos B - \sin A\cdot(-\sin B)$Using even/odd properties: $\cos(-B)=\cos B$, $\sin(-B)=-\sin B$
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4$= \cos A\cos B + \sin A\sin B$
Result (b)
$$\cos(A-B) = \cos A\cos B + \sin A\sin B$$
§5 Derivation of $\sin(A+B)$ and $\sin(A-B)$
We use the co-function identity: $\sin\theta = \cos(90° - \theta)$.
Derivation of $\sin(A+B)$:
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1$\sin(A+B) = \cos\bigl(90° - (A+B)\bigr)$Co-function identity: $\sin\theta = \cos(90°-\theta)$
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2$= \cos\bigl((90°-A) - B\bigr)$Rewriting: $90°-(A+B) = (90°-A)-B$
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3$= \cos(90°-A)\cdot\cos B + \sin(90°-A)\cdot\sin B$Apply formula (b): $\cos(X-Y)=\cos X\cos Y+\sin X\sin Y$, with $X=90°-A$, $Y=B$
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4$= \sin A\cdot\cos B + \cos A\cdot\sin B$Co-function identities: $\cos(90°-A)=\sin A$ and $\sin(90°-A)=\cos A$
Result (c)
$$\sin(A+B) = \sin A\cos B + \cos A\sin B$$
Derivation of $\sin(A-B)$:
Replace $B$ by $-B$ in formula (c):
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1$\sin(A-B) = \sin(A+(-B))$
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2$= \sin A\cdot\cos(-B) + \cos A\cdot\sin(-B)$Apply formula (c) with $B$ replaced by $-B$
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3$= \sin A\cdot\cos B + \cos A\cdot(-\sin B)$$\cos(-B)=\cos B$ and $\sin(-B)=-\sin B$
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4$= \sin A\cos B - \cos A\sin B$
Result (d)
$$\sin(A-B) = \sin A\cos B - \cos A\sin B$$
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§6 Derivation of $\tan(A+B)$ and $\tan(A-B)$
Derivation of $\tan(A+B)$:
Use the definition $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$:
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1$\tan(A+B) = \dfrac{\sin(A+B)}{\cos(A+B)}$Definition of tangent
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2$= \dfrac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}$Substituting results (c) and (a)
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3Divide numerator and denominator by $\cos A\cos B$: $$= \dfrac{\dfrac{\sin A\cos B}{\cos A\cos B} + \dfrac{\cos A\sin B}{\cos A\cos B}}{\dfrac{\cos A\cos B}{\cos A\cos B} - \dfrac{\sin A\sin B}{\cos A\cos B}}$$Dividing every term by $\cos A\cos B$
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4$= \dfrac{\dfrac{\sin A}{\cos A} + \dfrac{\sin B}{\cos B}}{1 - \dfrac{\sin A}{\cos A}\cdot\dfrac{\sin B}{\cos B}}$Simplifying each fraction
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5$= \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$$\dfrac{\sin\theta}{\cos\theta}=\tan\theta$
Result (e)
$$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$$
Derivation of $\tan(A-B)$:
Replace $B$ by $-B$ in formula (e):
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1$\tan(A-B) = \tan(A+(-B))$
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2$= \dfrac{\tan A + \tan(-B)}{1 - \tan A\cdot\tan(-B)}$Apply formula (e) with $B$ replaced by $-B$
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3$= \dfrac{\tan A + (-\tan B)}{1 - \tan A\cdot(-\tan B)}$$\tan(-B) = -\tan B$ (tangent is an odd function)
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4$= \dfrac{\tan A - \tan B}{1 + \tan A\tan B}$
Result (f)
$$\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$$
Page 5 of 5
§7 Derivation of $\cot(A+B)$ and $\cot(A-B)$
Derivation of $\cot(A+B)$:
Use $\cot\theta = \dfrac{1}{\tan\theta}$ and start from $\tan(A+B) = \dfrac{\tan A+\tan B}{1-\tan A\tan B}$.
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1$\dfrac{1}{\cot(A+B)} = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$$\tan(A+B) = \dfrac{1}{\cot(A+B)}$
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2Replace $\tan A = \dfrac{1}{\cot A}$ and $\tan B = \dfrac{1}{\cot B}$: $$\frac{1}{\cot(A+B)} = \dfrac{\dfrac{1}{\cot A}+\dfrac{1}{\cot B}}{1 - \dfrac{1}{\cot A}\cdot\dfrac{1}{\cot B}}$$$\tan = \frac{1}{\cot}$
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3Combine the fractions in numerator and denominator: $$= \dfrac{\dfrac{\cot B + \cot A}{\cot A\cot B}}{\dfrac{\cot A\cot B - 1}{\cot A\cot B}}$$Adding fractions: $\frac{1}{\cot A}+\frac{1}{\cot B}=\frac{\cot B+\cot A}{\cot A\cot B}$; subtracting: $1-\frac{1}{\cot A \cot B}=\frac{\cot A\cot B-1}{\cot A\cot B}$
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4$= \dfrac{\cot B + \cot A}{\cot A\cot B - 1}$$\cot A\cot B$ cancels
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5Take the reciprocal of both sides: $$\cot(A+B) = \dfrac{\cot A\cot B - 1}{\cot B + \cot A}$$
Result (g)
$$\cot(A+B) = \frac{\cot A\cot B - 1}{\cot B + \cot A}$$
Derivation of $\cot(A-B)$:
Replace $B$ by $-B$ in formula (g):
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1$\cot(A-B) = \cot(A+(-B))$
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2$= \dfrac{\cot A\cdot\cot(-B) - 1}{\cot(-B) + \cot A}$Apply formula (g) with $B$ replaced by $-B$
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3$= \dfrac{\cot A\cdot(-\cot B) - 1}{(-\cot B) + \cot A}$$\cot(-B) = -\cot B$ (cotangent is an odd function)
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4$= \dfrac{-\cot A\cot B - 1}{\cot A - \cot B}$
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5$= \dfrac{-(\cot A\cot B + 1)}{\cot A - \cot B}$Factor out $-1$ from numerator
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6Multiply numerator and denominator by $-1$: $$= \dfrac{\cot A\cot B + 1}{\cot B - \cot A}$$
Result (h)
$$\cot(A-B) = \frac{\cot A\cot B + 1}{\cot B - \cot A}$$
§8 Summary of All Eight Compound Angle Formulae
Result (a)
$$\cos(A+B) = \cos A\cos B - \sin A\sin B$$
Result (b)
$$\cos(A-B) = \cos A\cos B + \sin A\sin B$$
Result (c)
$$\sin(A+B) = \sin A\cos B + \cos A\sin B$$
Result (d)
$$\sin(A-B) = \sin A\cos B - \cos A\sin B$$
Result (e)
$$\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A\tan B}$$
Result (f)
$$\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A\tan B}$$
Result (g)
$$\cot(A+B) = \frac{\cot A\cot B - 1}{\cot B + \cot A}$$
Result (h)
$$\cot(A-B) = \frac{\cot A\cot B + 1}{\cot B - \cot A}$$
💡 Memory Tips
- For $\cos$: the signs are opposite — $\cos(A+B)$ has a minus, $\cos(A-B)$ has a plus.
- For $\sin$: the signs match — $\sin(A+B)$ has plus, $\sin(A-B)$ has minus.
- For $\tan$ and $\cot$: the denominator sign is always opposite to the compound angle sign.
New Millennium Academy · Grade 10 Additional Mathematics · Compound Angle Formulae
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Course material curated by Mr. Nripendraswar Acharya