Trigonometric Addition Formulas

Simple Derivation Of Trigonometric Addition Formulas

Trigonometry, just the name of it, can make you feel overwhelmed. When the students have their first encounter with trigonometry in the 10th standard,  it comes with a massacre of your marks. It's ok to be scared of trigonometry when you don't know much about it. The concepts are difficult to understand at first, but with time, you will eventually get hold of them. Today we are going to help you with one of the fundamental problems called trigonometric addition formulas. 

We know it is a basic trigonometry problem. Still, if you don't grasp it properly, it will lead to a domino effect leading you not to understand other complex equations and formulas that are waiting for you in trigonometry. So if you are a little rusty in remembering formulas, we can help you remember them with the help of a solved theorem. As a result, you will have a better understanding of how the Sin (a+b) formula came into existence. 

Theorem Of Sin (a + b) Proof

See trigonometry, as we know, is a branch of mathematics that uses triangles as its base to solve any problem. In most cases, when we are trying to solve a trigonometry problem using Sin (a+b) formula, we take the right angle triangle. We are going to take a step by step approach in deriving this theorem to prove trigonometric addition formula.

Before we start, you need to know in Sin (a+b) we are using two angles (a) and (b) both of these angles are acute. Meaning ∠a+∠b < 90°. 

Step 1:- First, you need to make a line OX that needs to rotate on the O axis in an anti-clockwise direction. Here the starting point of the line and the initial position will make an acute angle, which is named ∠XOY = a.

Step 2:- Here, you need to make the line rotate a little more in the same anti-clockwise direction, starting from its initial position, which is OY. 

Now, once you stop the rotation, your line's new position will make an acute angle ∠YOZ = b. 

Step 3:- Up until now, things were easy, but here you need to be careful as there are lots of things happening at the same time. We are going to divide this step into sub-parts, so you don't get confused. 

In step 3, you need to make a point A on line OZ and draw a perpendicular to OX, and the perpendicular meets at point B on OX. 

In addition to this, you need to make a point C on OY and draw a perpendicular from point C to OX. the perpendicular will meet point D on OX. 

Now connect point A to Point C, and from point C draw another perpendicular but now on AB. This perpendicular will meet at point E on AB.

As a result of all this construction now we have ∠EAC = ∠ECO = ∠XOY = a  

This is how your final construction will look like. 

Step 4:- With the above construction done now, we have a right angle triangle BOA; thus, we can start deriving our theorem. 

Sin (a+b) = \[\frac{BA}{AO}\]

= \[\frac{BE+EA}{AO}\]

=\[\frac{DC}{AO}+\frac{EA}{AO}\]  (BE = DE)

= \[\frac{EA}{CA}\;.\;\frac{CA}{AO}\;.\;\frac{DC}{CO}\;.\;\frac{CO}{AO}\]  putting up the values. 

= cos ∠EAC sin b + sin a cos b (now we know ∠EAC = a)

As a result, sin (a+b) = sin a cos b + cos a sin b 

There you go, that’s your sin (a+b) proof.

Solved Example

Now we are going to show you one example of sin(a+b) where you find the answer using trigonometric addition formulas. 

Q) Find the Value of Sin 75° Using Trigonometric Ratios 30° and 45°.

Answer 

here we have sin 75°,

which we can break into sin (45°+30°), here we have a = 45°, and b = 30°.

Now using our sin (a+b) formula.

= Sin 45° . cos 30° + cos 45° . sin 30°

= \[\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac{1}{2}\]

=  \[\frac{\sqrt{3}+1}{2\sqrt{2}}\] your final answer. 

That’s what trigonometric sin addition formula was all about, and we hope now you understand the method and can derive its theorem on your own. Trigonometry is a tough nut to crack, but if you keep on hitting (metaphorically), eventually it will break and you will be able to solve its problems and attain better scores in your exam—best of luck.