Basic and rich Trigonometry concepts part 2, proof of compound angle functions

Rich Trigonometry concepts, Compound angles

basic and rich trigonometry concepts part 2 compound angle relations proofs top

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.

Basic and rich Trigonometry concepts part 2-1 compound angles proofs

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.


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