## Basic Trigonometry

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.

At first look we may not find much application of trigonometry in real world, but one must consider, the modern civilization is based on the sinusoidal frequencies that are nothing but trigonometric functions.

An interesting aspect of simplification of complex trigonometric expressions is its tightly intertwined relationship with Algebra. Overall, trigonometry is a fascinating subject.

Trigonometry starts by defining its basic trigonometric functions linking the angle and side measurements of a right angled triangle. Let us see how.

A right angled triangle is the basis of all of trigonometry, however complex it is made finally.

### Important basic elements

- Right angled triangle: $\triangle ABC$ with right angle at $\angle ABC$.
- One of the opposite angles: $\angle {\theta}$.
- Bottom side:
**$base = b$**. - Perpendicular to the base,
**$height=h$**. - Diagonal side,
**$hypotenuse=l$.**

### Core geometrical concept

#### Pythagoras theorem:

$ (hypotenuse)^2 = (base)^2 + (height)^2$

Or, $l^2 = b^2 + h^2 $.

### Basic trigonometric functions and relations

Trigonometry starts by defining **two basic functions** on $\angle \theta$, namely, **$\sin\theta$** and **$\cos\theta$**.

$ \sin\theta = \displaystyle\frac{height}{hypotenuse}=\frac{h}{l}$ and,

$\cos\theta = \displaystyle\frac{base}{hypotenuse} = \frac{b}{l} $.

These are the two most basic trigonometric functions defined from which all other functions can be derived.

#### Relation between $\sin\theta$ and $\cos\theta$:

$ \sin^2\theta + \cos^2\theta = \left(\displaystyle\frac{h}{l}\right)^2 + \left(\displaystyle\frac{b}{l}\right)^2 = \left(\displaystyle\frac{h^2 + b^2}{l^2}\right)$

Or, $\sin^2\theta + \cos^2\theta = 1 $.

This is the most important and highly used relationship or identity in Trigonometry.

In principle then, if one of the $\sin\theta$ or $\cos\theta$ is given the other can be derived by applying Pythagoras theorem. So, we can say,

Only one of $\sin\theta$ and $\cos\theta$ is the most basic trigonometric function.

By defining these functions,

A definitive measurable relation is established between an angle and the sides of a right triangle.

#### Other trigonometric functions

The **third most important function** other than $\sin\theta$ and $\cos\theta$ is the **function $\tan\theta$**.

$ \tan\theta = \displaystyle\frac{\sin\theta}{\cos\theta} = \displaystyle\frac{height}{base} = \displaystyle\frac{h}{b} $

This defines the third possible direct ratio relation between pairs of sides of the right triangle $\triangle ABC$. In $\tan\theta$, the hypotenuse is canceled out and only height and base remain in the ratio.

**All the three functions** on $\angle\theta$ - $\sin\theta$, $\cos\theta$ and $\tan\theta$ are **pure numbers** as these are ratios.

The **inverse of these three functions** gives rise to **next three basic trigonometric functions**, namely, **$cosec\theta$**, **$sec\theta$** and **$cot\theta$**.

$ cosec\theta=\displaystyle\frac{1}{\sin\theta}=\displaystyle\frac{l}{h}$,

$sec\theta=\displaystyle\frac{1}{\cos\theta}=\displaystyle\frac{l}{b}$,

$cot\theta=\displaystyle\frac{1}{\tan\theta}=\displaystyle\frac{b}{h} $.

#### Other important relations between trigonometric functions

Apart from the most important relation betwen $\sin\theta$ and $\cos\theta$, the other important relations between the trigonometric functions are:

$ sec^2\theta = 1 + \tan^2\theta$, and,

$cosec^2\theta = 1 + \cot^2\theta $.

Also,

$cosec^2\theta + sec^2\theta = cosec^2\theta\sec^2\theta $.

#### Derivation:

$ sec^2\theta - 1 = \displaystyle\frac{1}{\cos^2\theta} - 1 $

$\hspace{21mm}= \displaystyle\frac{1 - \cos^2\theta}{\cos^2\theta} $

$\hspace{21mm}= \displaystyle\frac{\sin^2\theta}{\cos^2\theta}$

$\hspace{21mm}= \tan^2\theta $

Or, $sec^2\theta = 1 + \tan^2\theta $.

$ cosec^2\theta - 1 = \displaystyle\frac{1}{\sin^2\theta} - 1$

$\hspace{25mm}= \displaystyle\frac{1 - \sin^2\theta}{\sin^2\theta}$

$\hspace{25mm}= \displaystyle\frac{\cos^2\theta}{\sin^2\theta}$

$\hspace{25mm}= \cot^2\theta$

Or, $cosec^2\theta = 1 + \cot^2\theta $.

Also,

$ cosec\theta = \displaystyle\frac{1}{sin\theta}$, and $sec\theta = \displaystyle\frac{1}{cos\theta}$

So, $cosec^2\theta + sec^2\theta = \displaystyle\frac{\cos^2\theta +\sin^2\theta}{\sin^2\theta\cos^2\theta}$

Or, $cosec^2\theta + sec^2\theta = cosec^2\theta\sec^2\theta $.

These three along with the other basic trigonometric function relationships are all identities and form the base of trigonometric expressions. Using these building blocks, very complex trigonometric expressions can be created.

### Values of Trigonometric functions for specific angles

As each of the basic trigonometric functions is a ratio, for each value of angle $\angle\theta$ each of the trigonometric functions has a specific numerical value. These values have practical utilities, but the whole set of values are not to be memorized by a student. We will see more of which value is to be memorized how.

The first set of values are exhaustive and are to be referred to but not to be memorized by heart. The second set is much smaller and it is a better option for memorization by heart. But in any case, these values can also be quickly evaluated.

#### Larger set of trigonometric values

This is a useful and exhaustive list of values of the trigonometric functions for specific angles that are mostly used.

As $\tan\theta$ and other three functions can be derived from values of $\sin\theta$ and $\cos\theta$ easily not all these values are needed to be memorized. If we memorize only the values of $sin\theta$ and $cos\theta$ for specific angles $\theta$, rest of the values can be derived.

#### Smaller sets of trigonometric values

As mentioned above, from this small set of values of $\sin\theta$ and $cos\theta$, values of rest of the four functions can be easily derived.

One of the most distinctive characteristics of the value range of $sin\theta$ and $cos\theta$ is, both the functions change from $-1$ to $0$ to $+1$. **Absolute value of any of these two can never exceed 1.**

#### Minimum memorization of trigonometric values:

** Ultimately, if you remember: **

Value of $\sin\theta$ increases from $0$ to $1$ for angle $0^0$ to $90^0$, and,

- At $45^0$ values of both $\sin\theta$ and $\cos\theta$ are same and is $\displaystyle\frac{1}{\sqrt{2}}$.
- The values of $\sin\theta$ and $\cos\theta$ for angles $30^0$ and $60^0$ are either of $\displaystyle\frac{1}{2}$ and $\displaystyle\frac{\sqrt{3}}{2}$.
- Finally, $\sin\theta=\displaystyle\frac{1}{2}$ for $\theta=30^0$,

you will be able to derive all other values of the other trigonometric functions.

This approach reduces the memory load and increases accuracy of computation.

#### Still easier method to memorize values of $sin\theta$

But in spite of this simple situation of memorizing values of $sin\theta$ or $cos\theta$, we still sometimes mixed up the values of $sin30^0$ and $sin60^0$. To get over this slightly confusing situation of memorizing, we have devised a method which is concept based and thus is nearly fail-proof.

If you find it difficult to remember the values of $sin\theta$ and $cos\theta$ for various values of $\theta$, you just have to remember the following picture depicting variation of $sin\theta$ and $cos\theta$ for various values of $\theta$.

From the figure it can never be forgotten that $cos\theta=1$ when $\theta=0^0$ and its value reduces to $0$ when value of $\theta$ goes on increasing to $90^0$. In between there are two significant values of $cos\theta$, namely $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ at $\theta$ values either $30^0$ or $60^0$. As $\frac{1}{2}$ is smaller than $\frac{\sqrt{3}}{2}$, and also as value of $cos\theta$ reduces from 1 to 0, it follows that $cos60^0=\frac{1}{2}$.

This is direct application of the * Less facts and more procedure approach* or strategy.

#### Value change with change of angle by $\pi/2$ or $\pi$ or by negative sign:

When the angle of $\theta$ is changed by $\pi/2$ or by $\pi$ the values of the trigonometric functions change in simple ways.

**Change by $\pi/2$:**

$\sin\left(\displaystyle\frac{\pi}{2} - \theta\right) = \cos\theta$, and $\sin\left(\displaystyle\frac{\pi}{2} + \theta\right) = \cos\theta$

$\cos\left(\displaystyle\frac{\pi}{2} - \theta\right) = \sin\theta$, and $\cos\left(\displaystyle\frac{\pi}{2} + \theta\right) = -\sin\theta$.

**Change by $\pi$:**

$\sin\left({\pi} - \theta\right) = \sin\theta$, and $\sin\left({\pi} + \theta\right) = -\sin\theta$

$\cos\left({\pi} - \theta\right) = -\cos\theta$, and $\cos\left({\pi} + \theta\right) = -\cos\theta$.

**Change by sign:**

$\sin\left(-{\theta}\right) = -\sin\theta$, and $\cos\left(-{\theta}\right) = \cos\theta$.

#### Verification from Sin-Cos curve

You can verify all these relationships for change angle, by ${\pi}/2$, $\pi$ or by sign by comparing the actual values from the sin-cos curve above. These relationships follow the sin-cos curve pattern of change of value for change of angle.

### Rich concept - compound angles

Till now we have dealt with simple one angle for the trigonometric functions. That is the natural way to start as, finally you can't know the value of a $sin$ function, say, unless you provide the function with one single value.

But the way any subject gets more complex, trigonometry introduces in the same way, relationships between $sin$ $cos$ functions, for even the angle as a sum of two or more different angles.

For example **a standard relationship for a compound angle $(A + B)$ is,**

$\sin\left(A + B\right) = \sin A\cos B + \cos A\sin B$ ..........(F1)

By replacing $B$ with $-B$ you can always get,

$\sin\left(A - B\right) = \sin A\cos B - \cos A\sin B$

For $cos$ function similarly assume the following identity for compound sum angle,

$\cos\left(A + B\right) = \cos A\cos B - \sin A\sin B$ .............(F2)

Replacing $B$ again by $-B$ you would get,

$\cos\left(A - B\right) = \cos A \cos B + \sin A \sin B$.

Using the expressions of $\sin\left(A + B\right)$ and $\cos\left(A + B\right)$, the following compound angle relation for $\tan\left(A+B\right)$ can easily be derived.

$\tan\left(A+B\right)=\displaystyle\frac{\tan A+\tan B}{1-\tan A.{\tan B}}$.

For detailed proof of compound angle relations you may refer to **Basic and rich Trigonometry concepts part 2, proof of compound angle relations.**

### Rich concept - further derivation from compound angles

**Adding the two $sin$ relationships for compound angles** you will get,

$\sin(A + B) + \sin(A - B) = 2\sin A \cos B$, the $\cos A \sin B$s cancel out.

Or, $2\sin A \cos B = \sin(A + B) + \sin(A - B)$. ........(F3)

**Similarly adding two $cos$ compound angle relationships** you would get,

$2\cos A \cos B = \cos\left(A + B\right) + \cos\left(A - B\right)$ .........(F4)

**Derived formula by subtraction of two $sin$ compound angle relationships**

By subtraction from two $sin$ compound angle relationships,

$2\cos A\sin B = \sin\left(A + B\right) - \sin\left(A - B\right)$ ............(F5)

**Derived formula by subtraction from two $cos$ compound angle relationships**

$2\sin A\sin B = \cos\left(A - B\right) - \cos\left(A + B\right)$ ..............(F6)

#### Derived formulas for submultiple angles

Submultiple angle means fraction of an angle. For example angle $\displaystyle\frac{A}{2}$ is a submultiple angle for angle $A$. We are now onto derivation of formulas involving submultiple angles.

With the arsenal of formulas that we have developed by now it is easy to go into submuliples. Let us see how.

**Formula involving submultiple angle derived from formula $F3$**

In formula $F3$ if you substitute $\displaystyle\frac{A + B}{2}$ for $A$ and $\displaystyle\frac{A - B}{2}$ for $B$, you will get,

$\sin A + \sin B = 2\sin\left(\displaystyle\frac{A + B}{2}\right)\cos\left(\displaystyle\frac{A - B}{2}\right)$

**Formula involving submultiple angle derived from formula $F4$**

In formula $F4$ if you substitute $\displaystyle\frac{A + B}{2}$ for $A$ and $\displaystyle\frac{A - B}{2}$ for $B$, you will get,

$\cos A + \cos B = 2\cos\left(\displaystyle\frac{A + B}{2}\right)\cos\left(\displaystyle\frac{A - B}{2}\right)$

**Formula involving submultiple angle derived from formula $F5$**

In formula $F5$ if you substitute $\displaystyle\frac{A + B}{2}$ for $A$ and $\displaystyle\frac{A - B}{2}$ for $B$, you will get,

$\sin A - \sin B = 2\cos\left(\displaystyle\frac{A + B}{2}\right)\sin\left(\displaystyle\frac{A - B}{2}\right)$

**Formula involving submultiple angle derived from formula $F6$**

In formula $F6$ if you substitute $\displaystyle\frac{A + B}{2}$ for $A$ and $\displaystyle\frac{A - B}{2}$ for $B$, you will get,

$\cos A - \cos B = 2\sin\left(\displaystyle\frac{A + B}{2}\right)\sin\left(\displaystyle\frac{B - A}{2}\right)$, **alert:** *beware of the sign change.*

#### Derived multiple angles - 2A - substitute $A$ for $B$

$\sin 2A = 2\sin A\cos A$

$\cos 2A = \cos^2 A - \sin^2 A$

#### Derived multiple angles - 3A - substitute $2A$ for $B$

$\sin(A + 2A) = \sin A\cos 2A + \cos A\sin 2A$

Or, $\sin 3A = \sin A(\cos^2 A - \sin^2 A) + \cos A(2\sin A\cos A)$

$=3\sin A\cos^2A - \sin^3 A$

$=3\sin A(1 - \sin^2A) - \sin^3 A$.

So, finally,

$\sin 3A = 3\sin A - 4\sin^3 A$.

Similarly,

$\cos 3A = 4\cos^3 A - 3\cos A$.

This way you can derive any multiple or submultiples.

An **example of half-angle,**

$ \sin A = 2\sin \displaystyle\frac{A}{2}\cos \displaystyle\frac{A}{2} $.

**To add further complexity levels,** you can **consider changes in submultiples or multiples** *in the same way as the original singular angles.*

For all *compound angle, submultiple angle and multiple angle relationships then the basic relationships are only,*

- Three sets of
**change of angle relationships**that you can derive from the $sin-cos$ curve, - One
**compound angle $sin$ relationship $F1$**, and - One
**compound angle $cos$ relationship $F2$**.

**Note:** In SSC CGL and most of the competitive tests, Compound angles and multiple submultiple angles are generally not included, but occasionally knowledge of compound, multiple and submultiple angle functions help to solve difficult geometric and trigonometric problems elegantly. There is no harm in understanding and remembering primarily the compound angle functions from which other functions can be derived easily.

#### Law of Sine:

In a triangle $\triangle ABC$ with three sides as $BC=a$, $AC=b$, $AB=c$ and three angles $\angle A = \angle BAC$, $\angle B = \angle ABC$ and $\angle C = \angle ACB$, according to the Sine law,

$ \displaystyle\frac{\sin A}{A} = \displaystyle\frac{\sin B}{B} = \displaystyle\frac{\sin C}{C} = \text{Constant} $

### Application of Trigonometry - heights and distances

In this problem area, an object stands upright perpendicularly on the surface and from a distance the object is observed. If the angle of elevation is known along with one of the height or distance to the object, the other can be known.

This is one of the important application areas of Trigonometry.

#### Angle of elevation

When the observer looks up to the top of a tall object, the angle made by the horizontal line through point of observation and the line from the top of the object to the point of observation is the **angle of elevation**.

We have here, $\displaystyle\frac{h}{d}=\tan\theta$.

#### Angle of depression

When an observer looks down from the top of a tall object, we get **angle of depression.**

**Problem example**** 1:**

From a point $A$ on the bridge over a river, the angles of depression of the banks on the opposite sides of the river are $30^0$ and $45^0$ respectively. If the bridge is at a height of 9m above the river surface, find how wide is the river.

Width of the river is $DC$. In triangle $\triangle ABC$, $\tan 30^0=\displaystyle\frac{AB}{BC}$.

So, $BC=\displaystyle\frac{9}{\tan 30^0} = 9\sqrt{3} m$.

In $\triangle ABD$,

$\tan 45^0 = \displaystyle\frac{AB}{BD}$

$AB = BD$

$BD = 9m$

$DC = DB + BC = 9 + 9\sqrt{3} = 9(\sqrt{3} + 1)=24.588m$.

### Questions, Solutions and other resources on Trigonometry

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

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

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

**SSC CGL level Question Set 19, Trigonometry**

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

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

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

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

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

**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**