## Rich Trigonometry concepts, Compound angles

Trigonometry first established relationships between the angle and side measurements of a right angled triangle by defining trigonometric functions and thus extended the reach of mathematics wonderfully. Like any other branch of mathematics, from the basic trigonometric function definitions more advanced function relations are built stage by stage. Here we will deal with compound angle trigonometric function relations in more details.

You may refer to * Basic and rich Trigonometry concepts and its applications* for the foundational basic and rich trigonometry concepts. The rich concepts discussed here rest on the concepts introduced in the earlier discussion mentioned.

### Proof of relations for $\sin (A+B)$, $\cos (A+B)$ and $\tan (A+B)$

The following figure will help the solution process.

In the explanation, instead of $\angle A$ and $\angle B$ we will be using $\angle \alpha$ and $\angle \theta$ respectively.

In the figure, $\triangle APB$, $\triangle PFC$, $\triangle AFD$, $\triangle DEF$ and $\triangle AEF$ are right triangles where $EF||PBC$ and $EF=BC$ and $BE=CF$.

AS $EF||PB$, $\angle CPF=\angle EFD=\alpha$. Again in right $\triangle EFD$, $\angle \alpha +\angle x=90^0$ and in right $\triangle AFD$, $\angle x +\angle DAF=90^0$ so that, $\angle DAF=\angle \alpha$.

With this background work we are now ready to derive the expanded relationship of $\sin (\alpha +\theta)$.

$\sin (\alpha + \theta) = \displaystyle\frac{AB}{AP}$

$=\displaystyle\frac{AE+BE}{AP}$

$=\displaystyle\frac{CF}{AP}+ \displaystyle\frac{AE}{AP}$

$=\displaystyle\frac{CF}{PF}.\displaystyle\frac{PF}{AP}+\displaystyle\frac{AE}{AF}.\displaystyle\frac{AF}{AP}$

$=\sin \alpha.{\cos \theta} + \cos \alpha.{\sin \theta}$.

#### Proof of $\cos (A+B)$

Using the same figure above we have,

$\cos (\alpha +\theta)=\displaystyle\frac{PB}{AP}$

$=\displaystyle\frac{PC-BC}{AP}$

$=\displaystyle\frac{PC}{AP}- \displaystyle\frac{EF}{AP}$

$=\displaystyle\frac{PC}{PF}.\displaystyle\frac{PF}{AP}-\displaystyle\frac{EF}{AF}.\displaystyle\frac{AF}{AP}$

$=\cos \alpha.{\cos \theta} - \sin \alpha.{\sin \theta}$.

#### Proof of $\tan (A+B)$

Instead of deriving the relationship of $\tan (\alpha +\theta)$, using the results of $\sin (\alpha +\theta)$ and $\cos (\alpha +\theta)$ we will prove the relation,

$\tan (\alpha+\theta)=\displaystyle\frac{\tan \alpha +\tan \theta}{1-\tan \alpha.{\tan \theta}}$.

$\tan (\alpha+\theta)=\displaystyle\frac{\sin (\alpha +\theta)}{\cos (\alpha +\theta)}$

$=\displaystyle\frac{\sin \alpha.{\cos \theta} + \cos \alpha.{\sin \theta}}{\cos \alpha.{\cos \theta} - \sin \alpha.{\sin \theta}}$.

Taking out the factor $\cos \alpha.{cos \theta}$ from the numerator and denominaror of the RHS of the above relation we have,

$\tan (\alpha+\theta)=\displaystyle\frac{\tan \alpha +\tan \theta}{1-\tan \alpha.{\tan \theta}}$.

We need to understand the derivation of primarily $\sin (A +B)$ as derivation of $\cos (A + B)$ takes the same approach and from $\sin (A + B)$ and $\cos (A + B)$ the expression of $\tan (A +B)$ can easily be derived.

**Note:** Usually you won't need to derive the compound angle relations. Nevertheless understanding the proof enhances your belief on the important compound angle relationships that are used at various stages of Trigonometric and Geometric problem solving.

### Resources on Trigonometry and related topics

You may refer to our useful resources on Trigonometry and other related topics especially algebra.

### Tutorials on Trigonometry

**Basic and rich concepts in Trigonometry and its applications**

**Basic and Rich Concepts in Trigonometry part 2, proof of compound angle functions**

**Trigonometry concepts part 3, maxima (or minima) of Trigonometric expressions**

### General guidelines for success in SSC CGL

**7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests**

### Efficient problem solving in Trigonometry

**How to solve a School Math problem in a few direct steps, Trigonometry 5**

**How to solve difficult SSC CGL level School math problems in a few quick steps, Trigonometry 5**

**How to solve School Math problem in a few steps and in Many Ways, Trigonometry 4**

**How to solve a School Math problem in a few simple steps, Trigonometry 3**

**How to solve difficult SSC CGL level School math problems in a few simple steps, Trigonometry 4**

**How to solve difficult SSC CGL level School math problems in a few simple steps, Trigonometry 3**

**How to solve School math problems in a few simple steps, Trigonometry 2**

**How to solve School math problems in a few simple steps, Trigonometry 1**

**A note on usability:** The *Efficient math problem solving* sessions on **School maths** are **equally usable for SSC CGL aspirants**, as firstly, the "Prove the identity" problems can easily be converted to a MCQ type question, and secondly, the same set of problem solving reasoning and techniques have been used for any efficient Trigonometry problem solving.

### SSC CGL question and solution sets on Trigonometry

**SSC CGL Tier II level Solution Set 7 on Trigonometry 1**

**SSC CGL Tier II level Question Set 7 on Trigonometry 1**

**SSC CGL level Solution Set 40 on Trigonometry 4**

**SSC CGL level Question Set 40 on Trigonometry 4**

**SSC CGL level Solution Set 19 on Trigonometry**

**SSC CGL level Question set 19 on Trigonometry**

**SSC CGL level Solution Set 16 on Trigonometry**

**SSC CGL level Question Set 16 on Trigonometry**

**SSC CGL level Question Set 2 on Trigonometry**

**SSC CGL level Solution Set 2 on Trigonometry**

### Algebraic concepts

**Basic and rich Algebraic concepts for elegant solutions of SSC CGL problems**

**More rich algebraic concepts and techniques for elega****n****t solutions of SSC CGL problems**