Manipulating the Angle Addition Formulae

Maths: Edexcel A Level Pure Maths, topic 5.4.4

We can prove other angle addition formulae by manipulating what we have already for $\sin(A+B)$ and $\cos(A+B)$ .

We can find the identity for $\sin(A-B)$ by substituting $B \rightarrow-B$ into the identity for $\sin(A+B)$ :
$\sin(A+(-B)) \equiv \sin A\cos(- B) + \cos A\sin (-B)$
$\sin(A-B) \equiv \sin A\cos B - \cos A\sin B$
Where we have used $\cos(-B) = \cos B$ and $\sin(-B) = -\sin B$ .

We can find the identity for $\cos(A-B)$ by substituting $B \rightarrow-B$ into the identity for $\cos(A+B)$ :
$\cos(A+(-B)) \equiv \cos A\cos(- B) - \sin A\sin (-B)$
$\cos(A-B) \equiv \cos A\cos B +\sin A\sin B$
Where we have used $\cos(-B) = \cos B$ and $\sin(-B) = -\sin B$ .

To find the addition formula for tangent, we
$\tan(A+B) \equiv \frac{\sin(A+B)}{\cos(A+B)}$
Substituting in our angle addition formulas, we get:
$\tan(A+B) \equiv \frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B -\sin A\sin B}$
Dividing both the numerator and denominator by a factor of $\cos A\cos B$ gives:
$\tan(A+B) \equiv \frac{\small \frac{\sin A\cos B}{\cos A\cos B} + \frac{\cos A\sin B}{\cos A\cos B}}{\small \frac{\cos A\cos B}{\cos A\cos B} - \frac{\sin A \sin B}{\cos A \cos B}}$

Cancelling out common factors gives:
$\tan(A+B) \equiv \frac{\small \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\small 1 - \frac{\sin A \sin B}{\cos A \cos B}}$
Finally, we can simplify this as:
$\tan(A+B)\equiv \frac{\tan A+\tan B}{1 - \tan A \tan B}$

We can find the identity for $\tan(A-B)$ by substituting $B \rightarrow-B$ into the identity for $\tan(A+B)$ :
$\tan(A+(-B))\equiv \frac{\tan A+\tan (-B)}{1 - \tan A \tan(- B)}$
$\tan(A-B) \equiv \frac{\tan A-\tan B}{1 + \tan A\tan B}$
Where we have used $\tan(-B) = -\tan B$ .

Test yourself on Manipulating the Angle Addition Formulae