Compound Angle Identities
Compound angle identities are trigonometric formulas that express the sine, cosine, or tangent of the sum or difference of two angles in terms of the trigonometric functions of the individual angles.
Sine Compound Angle Identities
The sine compound angle identity is:
$$\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta$$
Let's calculate $\sin(75°)$ using the compound angle identity:
We can write 75° as 45° + 30°. Then:
$\sin(75°) = \sin(45° + 30°) = \sin(45°)\cos(30°) + \cos(45°)\sin(30°)$
$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$
Cosine Compound Angle Identities
The cosine compound angle identity is:
$$\cos(\alpha\pm\beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta$$
A sneaky part of the cosine compound angle formula is the $\mp$ sign on the right-hand side. This means that if you have $\cos(\alpha + \beta)$, you have to flip the sign to turn it into $\cos\alpha\cos\beta - \sin\alpha\sin\beta$, and vice versa for $\cos(\alpha - \beta)$.
Tangent Compound Angle Identities
We can derive the compound tangent identity from the compound sine and cosine identities.
First, we know that:
$$\tan(\alpha \pm \beta) = \frac{\sin(\alpha \pm \beta)}{\cos(\alpha \pm \beta)}$$
We can therefore substitute the identities for compound sine and cosine:
$$\tan(\alpha \pm \beta) = \frac{\sin\alpha\cos\beta \pm \cos\alpha\sin\beta}{\cos\alpha\cos\beta \mp \sin\alpha\sin\beta}$$
Dividing both the numerator and denominator by $\cos\alpha\cos\beta$, we get the compound tangent identity:
$$\tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta}$$
Remember, the denominator in the compound tangent formula has $\mp$ instead of $\pm$, meaning you have to swap the sign from $\tan(\alpha\pm\beta)$.
Derivation of Double Angle Identities
Double angle identities can be derived from compound angle identities by setting $B = A$ in the compound angle formulas.
Double Angle Identity for Sine
Starting with $\sin(A + B) = \sin A \cos B + \cos A \sin B$, let $B = A$:
$\sin(A + A) = \sin A \cos A + \cos A \sin A$
Therefore, $\sin(2A) = 2\sin A \cos A$
Double Angle Identity for Cosine
Starting with $\cos(A + B) = \cos A \cos B - \sin A \sin B$, let $B = A$:
$\cos(A + A) = \cos A \cos A - \sin A \sin A$
Therefore, $\cos(2A) = \cos^2 A - \sin^2 A$
This can also be written as: $\cos(2A) = 2\cos^2 A - 1$ or $\cos(2A) = 1 - 2\sin^2 A$
The multiple forms of the cosine double angle identity are useful in different contexts. Choose the one that best fits your problem!
Double Angle Identity for Tangent
The double angle identity for tangent is:
$\tan(2A) = \frac{2\tan A}{1 - \tan^2 A}$
This can be derived from the tangent compound angle identity:
$\tan(A + A) = \frac{\tan A + \tan A}{1 - \tan A \tan A} = \frac{2\tan A}{1 - \tan^2 A}$
Let's use the double angle identity to find $\tan(60°)$:
We know that $\tan(30°) = \frac{1}{\sqrt{3}}$
Using the double angle identity:
$\tan(60°) = \tan(2 \cdot 30°) = \frac{2\tan(30°)}{1 - \tan^2(30°)} = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - (\frac{1}{\sqrt{3}})^2} = \frac{2\sqrt{3}}{3 - 1} = \sqrt{3}$
Link to De Moivre's Theorem
We can use de Moivre's theorem, from topic 1, to prove identities for $\sin n\theta$ and $\cos n\theta$.
When $n = 2$, De Moivre's theorem gives:
$(\cos \theta + i \sin \theta)^2 = \cos(2\theta) + i \sin(2\theta)$
Expanding the left side:
$\cos^2 \theta - \sin^2 \theta + i(2\sin \theta \cos \theta) = \cos(2\theta) + i \sin(2\theta)$
Separating the real and imaginary components implies:
$\cos(2\theta) = \cos^2 \theta - \sin^2 \theta$
$\sin(2\theta) = 2\sin \theta \cos \theta$
These are exactly the double angle identities we derived earlier.
Find identities using de Moivre's theorem for:
- $\sin3\theta$ and $\cos3\theta$
- $\sin\frac12\theta$ and $\cos\frac12\theta$
