# What is the cos and sin of (15pi/4)?

Trigonometry is a branch of standardized mathematics that deals with the relationship between lengths, heights, and angles. Trigonometry is the branch of mathematics that deals with the ratios and the relations among the sides and angles in a triangle. Using Trigonometry can calculate various measurements connected to a triangle. Some standard ratios are defined for the ease of calculation of some common problems related to the length and angles of the sides of a right-angled triangle.

### Trigonometric Ratios

A trigonometric ratio is the proportion of sides with either of the acute angles in the right-angled triangle. It can be defined as a simple trigonometric ratio in terms of sides of a right-angled triangle i.e. the hypotenuse, base side, and the perpendicular side. There are three simple trigonometric ratios wiz. sine, cosine, and tangent.

- Sine is the function that takes in the parameter an angle θ, which is either of the acute-angles in the right-angled triangles and is defined as the ratio of the length of the opposite side to the hypotenuse of the right-angled triangle. In technical terms, it can be written as:
**sin(θ) = opposite side/hypotenuse** - Cosine is the function that takes in the parameter an angle θ, which is either of the acute-angles in the right-angled triangles and is defined as the ratio of the length of the adjacent side to the hypotenuse of the right-angled triangle. In technical terms, it can be written as:
**cos(θ) = adjacent side / hypotenuse** - Tangent is the function that takes in the parameter an angle θ, which is either of the acute-angles in the right-angled triangles and is defined as the ratio of the length of the opposite side to the adjacent side of the right-angled triangle. In technical terms, it can be written as:
**tan(θ) = opposite side / adjacent side**

These trigonometric ratios are related to one another using some trigonometric identities and formulas,

- tan(θ) = sin(θ) / cos(θ)
- sin
^{2}(θ) + cos^{2}(θ) = 1

Each of the trigonometric ratios has other three derived trigonometric ratios which are deduced by taking the inverse of the respective ratios. The Other three Trigonometric ratios are Cosecant, Secant, and Cotangent used mathematically as cosec, sec, and cot. These are related to the Primary trigonometric ratios as follows,

- cosec(θ) = 1 / sin(θ)
- sec(θ) = 1 / cos(θ)
- cot(θ) = 1 / tan(θ) = cos(θ) / sin(θ)

Below are some of the identities related to the standard trigonometric ratios and the derived trigonometric ratios,

- tan
^{2}(θ) + 1 = sec^{2}(θ) - cot
^{2}(θ) + 1 = cosec^{2}(θ)

### Trigonometric Table

The values of each trigonometric angle are fixed and defined, but some very common angles and their values are introduced for making the calculations easy, below is the table for some common angles and the basic trigonometric ratios.

Ratio\Angle(θ) | 0° | 30° | 45° | 60° | 90° |

sin(θ) | 0 | 1/2 | 1/√2 | √3/2 | 1 |

cos(θ) | 1 | √3/2 | 1/√2 | 1/2 | 0 |

tan(θ) | 0 | 1/√3 | 1 | √3 | ∞ |

cosec(θ) | ∞ | 2 | √2 | 2/√3 | 1 |

sec(θ) | 1 | 2/√3 | √2 | 2 | ∞ |

cot(θ) | ∞ | √3 | 1 | 1/√3 | 0 |

There are also some other trigonometric ratios to apply beyond the right-angled triangles,

- sin(-θ) = – sin(θ)
- cos(-θ) = cos(θ)
- tan(-θ) = – tan(θ)

And, these relations are used in the Cartesian Cooordiante System, these will also be used in solving the problem statement given,

- sin(nπ/2 + θ) = cos(θ), sin(nπ/2 – θ) = cos(θ)
- cos(nπ/2 + θ) = -sin(θ), cos(nπ/2 – θ) = sin(θ)
- sin(nπ + θ) = -sin(θ), sin(nπ – θ) = sin(θ)
- cos(nπ + θ) = -cos(θ), cos(nπ – θ) = -cos(θ)
- sin(3nπ/2 + θ) = -cos(θ), sin(3nπ/2 – θ) = -cos(θ)
- cos(3nπ/2 + θ) = sin(θ), cos(3nπ/2 – θ) = -sin(θ)
- sin(2nπ + θ) = sin(θ), sin(2nπ – θ) = -sin(θ)
- cos(2nπ + θ) = cos(θ),cos(2nπ – θ) = cos(θ)

There are also some special trigonometric formula for tangent function,

- cos (A + B) = [cos(A) × cos(B)] – [sin(A) × sin(B)]
- cos (A – B) = [cos(A) × cos(B)] + [sin(A) × sin(B)]
- sin (A + B) = [sin(A) × cos(B)] + [sin(B) × cos(A)]
- sin (A – B) = [sin(A) × cos(B)] – [sin(B) × cos(A)]

### What is the cos and sin of (15pi/4)?

**Method 1:**

Make use of some Compound Angle formulae to calculate the value of cos(15π/4). Here make use of the following identity or formula,

cos (A + B) = [cos(A) × cos(B)] – [sin(A) × sin(B)]

sin (A + B) = [sin(A) × cos(B)] + [sin(B) × cos(A)]

**Solution:**

cos(15π/4)

Write (15π/4) as (2π + 7π/4) So,

cos(15π/4) = cos(2π + 7π/4)

cos (A + B) = [cos(A) × cos(B)] – [sin(A) × sin(B)]Here, A = 2π and B = 7π/4

cos(15π/4) = cos(2π + 7π/4)

= [cos(2π) × cos(7π/4)] – [sin(2π) × sin(7π/4)]

= [(1) × cos(2π – (π/4)) – [(0) × sin(2π – (π/4))]

= [cos(π/4)] – [0]

= 1/√2

sin (A + B) = [sin(A) × cos(B)] + [sin(B) × cos(A)]Here, A = 2π and B = 7π/4

sin(15π/4) = sin(2π + 7π/4)

= [sin(2π) × cos(7π/4)] + [cos(2π) × sin(7π/4)]

= [(0) × cos(2π – (π/4))] + [(1).sin(2π – (π/4))]

= [0] + [-sin(π/4)]

= -1/√2

Therefore,

sin(15π/4) = -1/√2

cos(15π/4) = 1/√2

**Method 2:**

Make use of simple trigonometric identity to calculate the value of cos(15π/4) and sin(15π/4). Here make use of the following identity,

sin(2π – θ) = – sin(θ)

cos(2π – θ) = cos(θ)

**Solution**:

(15π/4)

Write (15π/4) as (4π/ – π/4), So,

sin(15π/4) = cos(4π – π/4)

sin(2π – θ) = – sin(θ)Here, θ = π/4 = 45°

sin(15π/4) = sin(4π – π/4)

= sin [ (2(2π)) – (π/4) ]

= – sin (π/4)

= -1/√2

Similarly,

cos(2π – θ) = cos(θ)here, θ = π/4

cos(15π/4) = cos(4π – π/4)

= cos [ (2(2π)) – (π/4) ]

= cos (π/4)

= 1/√2

Therefore,

sin(15π/4) = -1/√2

cos(15π/4) = 1/√2

**Method 3**

Make use of trigonometric identities and Compound Angle Formulae to calculate the value of sin(15π/4) and cos(15π/4),

**Solution**:

cos(15π/4)

Write (15π/4) as (3π + 3π/4), So,

cos(15π/4) = cos(3π + 3π/4)

The compound angle formulae,

cos (A + B) = (cos(A) × cos(B)) – (sin(A) × sin(B)),Here, A = 3π and B = 3π/4

Therefore,

cos(15π/4) = cos( 3π + 3π/4)

= [ cos(3π).cos(3π/4) ] – [ sin(3π).sin(3π/4) ]

= [ (0).cos(π/2+(π/4)) ] – [ (-1).sin(π/2 + (π/4]) ]

= [ -sin(π/4).(0) ] – [ -cos(π/4) ]

= 0 – (-1/√2)

= 1/√2

Now,

sin (A + B) = [sin(A) × cos(B)] + [sin(B) × cos(A)]sin(15π/4) = sin(3π + 3π/4)

= [sin(3π) × cos(3π/4)] – [cos(3π) × sin(3π/4)]

= [(-1) × cos(π/2 + (π/4))] – [(0) × sin(π/2 + (π/4])]

= [-sin(π/4)] – [(0) × (cos(π/4)]

= (-1/√2) – (0)

= -1/√2

Therefore,

sin(15π/4) = -1/√2

cos(15π/4) = 1/√2

**So from the above methods, the value of cos(15π/4) and sin(15π/4) are calculated which are 1/√2 and -1/√2 respectively.**

### Sample Problems

**Question 1: Find the value of cos(3π/4)**

**Solution**:

Write (3π/4) as (π/2 + π/4) So,

cos(3π/4) = cos (π/2 + π/4)

cos(nπ/2 + θ) = – sin(θ)Here, θ = π/4

cos(3π/4) = – sin(π/4)

= – 1/√2

Therefore,

cos(3π/4) = -1/√2

**Question 2: Find the value of sin(19π/6)**

**Solution**:

Write (19π/6) as (3π + π/6) So,

sin(19π/6) = sin(3π + π/6)

sin(nπ + θ) = – sin(θ)Here, θ = π/6

sin(19π/6) = – sin(π/6)

= -1/2

= -0.5

Therefore,

sin(19π/6) = -0.5

**Question 3: Find cosec and sec of (7π/6)**

**Solution:**

Write (7π/6) as (π + π/6) Therefore,

sin(7π/6) = sin(π + π/6)

Since,

sin(nπ + θ) = – sin(θ)sin(7π/6) = sin(π + π/6)

= – sin(π/6)

= -1/2

Since,

cosec(θ) = 1 / sin(θ)Therefore,

cosec(7π/6) = 1/ sin(7π/6)

= 1 / (-1/2)

= -2

Now, cos(7π/6) = cos(π + π/6)

Since,

cos(nπ + θ) = – cos(θ)cos(7π/6) = cos(π + π/6]

= – cos(π/6)

= – √3/2

Since,

sec(θ) = 1/cos(θ)sec(7π/6) = 1 / cos(7π/6)

= 1 / (-√3/2)

= -2/√3

Thus,

cosec(7π/6) = -2

sec(7π/6) = -2/√3