### SubsectionB.2.5Identities — Adding Angles

We wish to explain the origins of the identity

\begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta). \end{align*}

A very geometric demonstration uses the figure below and an observation about areas. • The left-most figure shows two right-angled triangles with angles $\alpha$ and $\beta$ and both with hypotenuse length $1\text{.}$
• The next figure simply rearranges the triangles — translating and rotating the lower triangle so that it lies adjacent to the top of the upper triangle.
• Now scale the lower triangle by a factor of $q$ so that edges opposite the angles $\alpha$ and $\beta$ are flush. This means that $q \cos \beta = \cos \alpha\text{.}$ ie
\begin{align*} q &= \frac{\cos\alpha}{\cos\beta} \end{align*}
Now compute the areas of these (blue and red) triangles
\begin{align*} A_\text{red} &= \frac{1}{2} q^2 \sin\beta \cos \beta\\ A_\text{blue} &= \frac{1}{2} \sin \alpha \cos \alpha\\ \end{align*}

So twice the total area is

\begin{align*} 2 A_\text{total} &= \sin \alpha \cos \alpha + q^2 \sin\beta \cos \beta \end{align*}
• But we can also compute the total area using the rightmost triangle:
\begin{align*} 2 A_\text{total} &= q \sin(\alpha+\beta) \end{align*}

Since the total area must be the same no matter how we compute it we have

\begin{align*} q \sin(\alpha+\beta) &= \sin \alpha \cos \alpha + q^2 \sin\beta \cos \beta\\ \sin(\alpha+\beta) &= \frac{1}{q} \sin \alpha \cos \alpha + q \sin\beta \cos \beta\\ &= \frac{\cos \beta}{\cos \alpha} \sin \alpha \cos \alpha + \frac{\cos \alpha}{\cos \beta} \sin\beta \cos \beta\\ &= \sin \alpha \cos \beta + \cos \alpha \sin\beta \end{align*}

as required.

We can obtain the angle addition formula for cosine by substituting $\alpha \mapsto \pi/2-\alpha$ and $\beta \mapsto -\beta$ into our sine formula:

\begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) & \text{becomes}\\ \underbrace{\sin(\pi/2-\alpha-\beta)}_{\cos(\alpha+\beta)} &= \underbrace{\sin(\pi/2-\alpha)}_{\cos(\alpha)}\cos(-\beta) + \underbrace{\cos(\pi/2-\alpha)}_{\sin(\alpha)}\sin(-\beta)\\ \cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \end{align*}

where we have used $\sin(\pi/2-\theta)=\cos(\theta)$ and $\cos(\pi/2-\theta)=\sin(\theta)\text{.}$

It is then a small step to the formulas for the difference of angles. From the relation

\begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \end{align*}

we can substitute $\beta \mapsto -\beta$ and so obtain

\begin{align*} \sin(\alpha - \beta) &= \sin(\alpha)\cos(-\beta) + \cos(\alpha)\sin(-\beta)\\ &= \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) \end{align*}

The formula for cosine can be obtained in a similar manner. To summarise

\begin{align*} \sin(\alpha \pm \beta) &= \sin(\alpha)\cos(\beta) \pm \cos(\alpha)\sin(\beta)\\ \cos(\alpha \pm \beta) &= \cos(\alpha)\cos(\beta) \mp \sin(\alpha)\sin(\beta) \end{align*}

The formulas for tangent are a bit more work, but

\begin{align*} \tan(\alpha + \beta) &= \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}\\ &= \frac{\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) }\\ &= \frac{\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) } \cdot \frac{\sec(\alpha) \sec(\beta)}{\sec(\alpha) \sec(\beta)}\\ &= \frac{\sin(\alpha)\sec(\alpha) + \sin(\beta)\sec(\beta)}{1 - \sin(\alpha)\sec(\alpha)\sin(\beta)\sec(\beta) }\\ &= \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta) }\\ \end{align*}

and similarly we get

\begin{align*} \tan(\alpha - \beta) &= \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta) } \end{align*}