Cosine of a compound angle that formed by the addition of two angles of a right angled triangle can be expressed in terms of trigonometric ratios cosine and sine with the same two angles.

For example, $A$ and $B$ are two angles and $A+B$ is the compound angle of the right angled triangle. $\cos(A+B)$ is called cosine of compound angle and it can be expressed in terms of cosine and sine with angles $A$ and $B$.

$\cos (A+B) \,\,=\,\, \cos A . \cos B \,\, – \,\, \sin A . \sin B$

A right angled triangle is constructed with a compound angle to get the expansion of cosine of the compound angle in mathematical form.

It is highly recommended to study the properties of this triangle from our step by step construction procedure of the triangle to understand the derivation of cosine of the angle.

1

$$\cos (A+B) = \frac{MS}{MQ}$$

The side $\overline{MN}$ is split into two sides $\overline{MS}$ and $\overline{SN}$. So, the length of the $\overline{MS}$ can be obtained by subtracting length of $\overline{SN}$ from $\overline{MN}$.

$MN = MS + SN$

$\implies MS = MN \, – \, SN$

$$\therefore \,\, \cos (A+B) = \frac{MS}{MQ} = \frac{MN \, – \, SN}{MQ}$$

$$ \implies \cos (A+B) = \frac{MN}{MQ} \, – \, \frac{SN}{MQ}$$

2

$$\cos A = \frac{MN}{MP}$$

$\implies MN = MP \times \cos A$

Substitute the length of $\overline{MN}$ in right hand side of the $\cos (A+B)$.

$$\therefore \,\, \cos (A+B) = \frac{MP \times \cos A}{MQ} \, – \, \frac{SN}{MQ}$$

$$\implies \cos (A+B) = \frac{MP}{MQ} \times \cos A \,\, – \frac{SN}{MQ}$$

3

$$\cos B = \frac{MP}{MQ}$$

Replace ratio of lengths of sides $MP$ to $MQ$ by $\cos B$ in the expression of $\cos(A+B)$

$$\therefore \,\, \cos (A+B) = \cos B \times \cos A \,\, – \frac{SN}{MQ}$$

$$\implies \cos (A+B) = \cos A \cos B \,\, – \frac{SN}{MQ}$$

The length of the side $\overline{SN}$ is exactly equal to the length of the $\overline{UP}$. Therefore, $SN = UP$

$$\therefore \,\, \cos (A+B) = \cos A \cos B \,\, – \frac{UP}{MQ}$$

4

$$\sin A \,\,=\,\, \frac{UP}{QP}$$

$\implies UP \,\,=\,\, QP \times \sin A $

Now replace length of the $\overline{UP}$ by its value in the expression of the $\cos(A+B)$

$$\therefore \,\, \cos (A+B) \,\, = \,\, \cos A \cos B \,\,\,\, – \,\, \frac{QP \times \sin A}{MQ}$$

$$\implies \cos (A+B) \,\,=\,\, \cos A \cos B \,\,\,\, – \,\, \frac{QP}{MQ}\times \sin A$$

5

Once again consider right angled triangle $PMQ$.

$$\sin B \,\,=\,\, \frac{QP}{MQ}$$

Replace the ratio of lengths of the sides $\overline{QP}$ to $\overline{MQ}$ by $\sin B$ in $\cos (A+B)$ to obtain the require result.

$$\therefore \,\, \cos (A+B) \,\,=\,\, \cos A \cos B \,\,\, – \,\, \sin B \sin A$$

It can be written as follows

$$\cos (A+B) = \cos A . \cos B \,\,\, – \,\, \sin A . \sin B$$

The two angles which form a compound angle by addition can be represented by any two symbols but the expansion of the cosine of compound angle is same.

If two angles are $x$ and $y$, it is written as follows

$\cos (x+y) = \cos x \cos y \,-\, \sin x \sin y$

If two angles are $\alpha$ and $\beta$, it is written as follows

$\cos (\alpha + \beta) = \cos \alpha \cos \beta \,-\, \sin \alpha \sin \beta$