Trigonometry is one of the major parts of mathematics. In trigonometry, we study six predefined functions sine (sin), cosine (cos), tangent (tan), cosecant (cosec), secant (sec) and cotangent (cot). These functions relate the angles of a triangle to the sides of the triangle.

## Cosine Definition

If in a right triangle, the longest side is the hypotenuse, the side that is along the angle taken in consideration (let x) apart from the hypotenuse is the adjacent and the side opposite to the angle is the opposite, then we define

cos x = $\frac{adjacent}{hypotenuse}$

sin x = $\frac{opposite}{hypotenuse}$

We defined sin x as well here as we need it to explain the use of cosine function in tangent function. The tangent function is defined as

tan x = $\frac{sin x}{cos x}$

The domain of the cosine function is set of real numbers, R and the range is [-1, 1]. The period of the cosine is 2$\pi$

## Graph of Cosine

Some common values of cosine are

Cos 0$^o$ = 1

Cos 30$^o$= $\frac{(\sqrt3)}{2}$

Cos 45$^o$ = $\frac{1}{(\sqrt 2)}$

Cos 60$^o$= $\frac{1}{2}$

Cos 90$^o$ = 1

## Identities and Formulas Using Cosine

Few properties of trigonometric functions containing cosine are given below:
• Sin$^{2}$ x + cos$^{2}$ x = 1

• Cos 2x = cos$^{2}$ x - sin$^{2}$ x = 2cos$^{2}$ x - 1 = 1 - 2sin$^{2}$ x

• Cos (x + y) = cos x cos y - sin x sin y

• Cos (x - y) = cos x cos y + sin x sin y

• Sec x = $\frac{1}{cos x}$

• Cos $(\frac{\pi}{2}-x)$ = sin x

• Sin $(\frac{\pi}{2}- x)$ = cos x

• Cos (-x) = cos x

• Tan$^{2}$ x = $\frac{(1 - cos\ 2x)}{(1 + cos\ 2x)}$

• Sin x + sin y = 2 sin $\frac{(x + y)}{(2)}$ cos $\frac{(x - y)}{(2)}$

• Sin x - sin y = 2 sin $\frac{(x - y)}{(2)}$ cos $\frac{(x + y)}{(2)}$

• Cos x + cos y = 2 cos $\frac{(x + y)}{(2)}$ cos $\frac{(x - y)}{(2)}$

• Cos x - cos y = - 2 sin $\frac{(x + y)}{(2)}$ sin $\frac{(x - y)}{(2)}$

## Inverse Cosine

If y = cos x, then the inverse function of cosine is defined as x = cos$^{(-1)}$ y

Range of the inverse cosine function is [0, $\pi$] and the domain is [-1, 1]

Also cos (cos$^{-1}$ x) = x

## Inverse Cosine Identities

cos$^{-1}$ x = sin$^{-1}$ ($\sqrt {1 - x^{2}}$

sin$^{-1}$ x = cos$^{-1}$ ($\sqrt{1 - x^{2}}$

## Hyperbolic Cosine

We define hyperbolic cosine as

cosh x = $\frac{1}{2}$ (e$^{x}$ + e$^{-x}$)

The hyperbolic cosine function basically describes the shape of that of the hanging cable which is also known as the catenary.

It is important to note that cosh 0 = 1 and cosh (ln $\phi$) = $\frac{1}{2}$ ($\sqrt5$)

Here, $\phi$ is the golden ratio.

The hyperbolic cosine has Taylor series expansion that is,

Cosh z = 1 + $\frac{1}{2}$ x$^{2}$ + $\frac{1}{4}$! x$^{4}$ + $\frac{1}{6}$! x$^{6}$ + … = $\sum_{(i = 0)}^{\infty}$ [$\frac{ (x^{(2n)})}{ (2n)!}$]

The graph of the hyperbolic cosine is below.

## Law of Cosines

In a triangle ABC, AB = c, BC = a and AC = b and angles are A, B and C. then we make use of cosine law in order to determine if any side is missing or the angles if all sides are given

According to the cosine law we have

Cos A = $\frac{(b^{2} + c^{2} - a^{2})}{2bc}$

Cos B = $\frac{(a^{2} + c^{2} - b^{2})}{2ac}$

Cos C = $\frac{(a^{2} + b^{2} - c^{2})}{2ab}$

This law is very useful in determining angles of a triangle if all sides are known and hence in making judgment of the type of the triangle being acute, right or obtuse

## Examples on Cosine

Let us see some example on cosine function

Example 1:

Find cos 75$^o$

Solution:

cos 75$^o$

= cos (45$^o$ + 30$^o$)

= cos 45$^o$ cos 30$^o$ - sin 45$^o$ sin 30$^o$

= $\frac{1}{(\sqrt2)}$ x $\frac{(\sqrt3)}{2}$ - $\frac{1}{(\sqrt2)}$ x $\frac{1}{2}$

= $\frac{[(\sqrt3) - 1]}{[2 (\sqrt2)]}$

Example 2:

If the sides of the triangle are 3, 4, 5 respectively, then what type of triangle it is?

Solution:

Let a = 3, b = 4, c = 5

Cos A = $\frac{(b^{2} + c^{2} - a^{2})}{2bc}$

$\Rightarrow$ Cos A = $\frac{(16 + 25 - 9)}{40}$

$\Rightarrow$ Cos A = $\frac{4}{5}$

Cos B = $\frac{(a^{2} + c^{2} - b^{2})}{2ac}$

$\Rightarrow$ Cos B = $\frac{(9 + 25 - 16)}{30}$

$\Rightarrow$ Cos B = $\frac{3}{5}$

Then cos C = $\frac{(a^{2} + b^{2} - c^{2})}{2ab}$

$\Rightarrow$ Cos C = $\frac{(9 + 16 - 25)}{24}$

$\Rightarrow$ Cos C = 0

Cosine is equal to zero when the angle is 90?

$\Rightarrow$ Angle C = 90$^{\circ}$

Since one angle is a right angle, hence the triangle with given dimensions will be a right triangle.