IB Mathematics Analysis and Approaches AHL 3.10—compound Angle Identities Notes

Why sin(A + B) Is Not sinA + sinB

1. It is tempting to guess that $\sin(A+B) = \sin A + \sin B$, treating $\sin$ like a bracket you can multiply out. That guess is wrong.
2. Quick test with $A = B = 30^\circ$: $\sin 60^\circ = \tfrac{\sqrt3}{2} \approx 0.87$, but $\sin 30^\circ + \sin 30^\circ = 1$. The two are not equal, so a real formula is needed.
3. The compound angle identities are that formula. They express the sine, cosine, or tangent of a sum or difference of two angles using the ratios of the separate angles.
4. They are the workhorse of HL trigonometry: they generate the double-angle identities, $\tan 2\theta$, and the half-angle results, they let you find exact values like $\cos 15^\circ$, and they reappear in De Moivre's theorem and in calculus.

Deriving cos(A − B): The Master Identity

1. Rather than memorise six identities, derive just one carefully.
2. Everything else falls out of it with a sign flip or a co-function swap.
3. Place two points on the unit circle: $P = (\cos A, \sin A)$ at angle $A$, and $Q = (\cos B, \sin B)$ at angle $B$.
4. The angle between the radii $OP$ and $OQ$ is $A - B$.
5. Find the distance $PQ$ two ways. First, with the distance formula: $$PQ^2 = (\cos A - \cos B)^2 + (\sin A - \sin B)^2$$
6. Expand and use $\sin^2\theta + \cos^2\theta = 1$ on both angles: $$PQ^2 = 2 - 2(\cos A\cos B + \sin A\sin B)$$
7. Now find $PQ^2$ a second way, with the cosine rule in triangle $OPQ$, where $OP = OQ = 1$ and the included angle is $A - B$: $$PQ^2 = 1^2 + 1^2 - 2(1)(1)\cos(A - B) = 2 - 2\cos(A - B)$$
8. Both expressions equal $PQ^2$, so set them equal and cancel the $2$: $$\cos(A - B) = \cos A\cos B + \sin A\sin B$$

Getting The Other Five From cos(A − B)

Cosine Of A Sum

Replace $B$ with $-B$ in the master identity, and use that cosine is even ($\cos(-B)=\cos B$) and sine is odd ($\sin(-B)=-\sin B$):

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

Sine Of A Sum And Difference

1. Use the co-function link $\sin\theta = \cos\!\left(\tfrac{\pi}{2} - \theta\right)$ with $\theta = A + B$: $$\sin(A+B) = \cos\!\left(\tfrac{\pi}{2} - A - B\right) = \cos\!\left(\left(\tfrac{\pi}{2} - A\right) - B\right)$$
2. Apply the master identity to that difference, then convert back with $\cos\!\left(\tfrac{\pi}{2}-A\right)=\sin A$ and $\sin\!\left(\tfrac{\pi}{2}-A\right)=\cos A$: $$\sin(A + B) = \sin A\cos B + \cos A\sin B$$
3. Replace $B$ with $-B$ again to get the difference version: $$\sin(A - B) = \sin A\cos B - \cos A\sin B$$

Tangent Of A Sum And Difference

1. Start from $\tan(A \pm B) = \dfrac{\sin(A \pm B)}{\cos(A \pm B)}$ and substitute the identities just found: $$\tan(A \pm B) = \frac{\sin A\cos B \pm \cos A\sin B}{\cos A\cos B \mp \sin A\sin B}$$
2. Divide the top and bottom by $\cos A\cos B$; every term becomes a tangent: $$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$$
3. The full set of compound angle identities:

$$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$$

$$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$$

$$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$$

Finding Exact Values

Compound identities let you build angles like $15^\circ$ or $75^\circ$ out of the standard angles $30^\circ$, $45^\circ$, $60^\circ$.

Double Angle Identities: Set B = A

1. The most common special case of a compound angle is when the two angles are equal.
2. Setting $B = A$ turns each compound identity into a double angle identity.

Sine And Cosine

1. From $\sin(A + B) = \sin A\cos B + \cos A\sin B$ with $B = A$: $$\sin 2A = 2\sin A\cos A$$
2. From $\cos(A + B) = \cos A\cos B - \sin A\sin B$ with $B = A$: $$\cos 2A = \cos^2 A - \sin^2 A$$
3. Two more forms come from substituting $\sin^2 A = 1 - \cos^2 A$ or $\cos^2 A = 1 - \sin^2 A$: $$\cos 2A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$$

Tangent

From $\tan(A + B)$ with $B = A$: $$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$$

Using Double Angles In Proofs

A frequent HL task is to simplify an expression by spotting a double angle, then cancelling.

A Second Route: De Moivre's Theorem

1. The double angle identities also drop out of De Moivre's theorem (Topic 1), which is a useful cross-check and the standard way to reach $\sin 3\theta$ and $\cos 3\theta$.
2. With $n = 2$, De Moivre's theorem says: $$(\cos\theta + i\sin\theta)^2 = \cos 2\theta + i\sin 2\theta$$
3. Expand the left side and group real and imaginary parts: $$\cos^2\theta - \sin^2\theta + i\,(2\sin\theta\cos\theta) = \cos 2\theta + i\sin 2\theta$$
4. Match real parts and imaginary parts separately: $$\cos 2\theta = \cos^2\theta - \sin^2\theta, \qquad \sin 2\theta = 2\sin\theta\cos\theta$$
5. These are exactly the double angle identities from before, reached without the compound formulas.