# Sin Cos Formulas in Trigonometry with Examples

• Last Updated : 24 Jul, 2022

Trigonometry, as its name implies, is the study of triangles. It is an important branch of mathematics that studies the relationship between side lengths and angles of the right triangle and also aids in determining the missing side lengths or angles of a triangle. There are six trigonometric ratios or functions: sine, cosine, tangent, cosecant, secant, and cotangent, where cosecant, secant, and cotangent are the reciprocal functions of the other three functions, i.e., sine, cosine, and tangent, respectively. A trigonometric ratio is defined as the ratio of the side lengths of a right triangle. Trigonometry is employed in various fields in our daily life. It helps to determine the heights of hills or buildings. It is also used in fields like criminology, construction, physics, archaeology, marine engine engineering, etc.

### Formulae of six trigonometric ratios/functions

Let us consider a right-angled triangle XYZ, where ∠Y = 90°. Let the angle at vertex Z be θ. The side adjacent to “θ” is called the adjacent side, and the side opposite to “θ” is called the opposite side. A hypotenuse is a side opposite to the right angle or the longest side of a right angle.

• sin θ = Opposite side/Hypotenuse
• cos θ = Adjacent side/Hypotenuse
• tan θ = Opposite side/Adjacent side
• cosec θ = 1/sin θ = Hypotenuse/Opposite side
• sec θ = 1/ cos θ = Hypotenuse/Adjacent side
• cot θ = 1/ tan θ = Adjacent side/Opposite side

### Sine Formula

The sine of an angle in a right-angled triangle is the ratio of the length of the opposite side to the length of the hypotenuse to the given angle. A sine function is represented as “sin”.

sin θ = Opposite side/Hypotenuse

### Cosine Formula

The cosine of an angle in a right-angled triangle is the ratio of the length of the adjacent side to the length of the hypotenuse to the given angle. A cosine function is represented as “cos”.

### Some Basic Sine and Cosine Formulae

Sine and Cosine Functions in Quadrants

• The sine function is positive in the first and second quadrants and negative in the third and fourth quadrants.
• The cosine function is positive in the first and fourth quadrants and negative in the second and third quadrants.

The negative angle identity of the sine and cosine functions

• The sine of a negative angle is always equal to the negative sine of the angle.

sin (– θ) = – sin θ

• The cosine of a negative angle is always equal to the cosine of the angle.

cos (– θ) = cos θ

Relation between sine and cosine function

sin θ = cos (90° – θ)

Reciprocal functions of the sine and cosine functions

• A Cosecant function is the reciprocal function of the sine function.

cosec θ = 1/sin θ

• A Secant function is the reciprocal function of the cosine function.

sec θ = 1/cos θ

Pythagorean identity

sin2θ + cos2θ = 1

Periodic identities of the sine and cosine functions

sin (θ + 2nπ) = sin θ

cos (θ + 2nπ) = cos θ

Double Angle formulae for the sine and cosine functions

sin 2θ = 2 sin θ cos θ

cos 2θ = cos2θ – sin2θ = 2 cos2θ – 1 = 1 – 2 sin2θ

Half-angle identities for the sine and cosine functions

sin (θ/2) = ±√[(1 – cos θ)/2]

cos (θ/2) = ±√[(1 + cos θ)/2]

Triple angle identities for the sine and cosine functions

sin 3θ = 3 sin θ – 4 sin3θ

cos 3θ = 4cos3θ – 3 cos θ

Sum and difference formulas

• Sine function

sin (A + B) = sin A cos B + cos A sin B

sin (A – B) = sin A cos B – cos A sin B

• Cosine function

cos (A + B) =  cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

Law of sines or Sine Rule

The law of sines of sine rule is a trigonometric law that gives a relationship between the side lengths and angles of a triangle.

a/sin A = b/sin B = c/sin C

Where a, b, and c are the lengths of the three sides of the triangle ABC, and A, B, and C are the angles.

Law of cosines

The law of cosines of cosine rule is used to determine the missing or unknown angles or side lengths of a triangle.

a2 = b2 + c2 – 2bc cos A

b2 = c2 + a2 – 2ca cos B

c2 = a2 + b2 – 2ab cos C

Where a, b, and c are the lengths of the three sides of the triangle ABC, and A, B, and C are the angles.

### Problems based on Sine Formula and Cosine Formula

Problem 1: If cos α = 24/25, then find the value of sin α.

Solution:

Given,

cos α = 24/25

From the Pythagorean identities we have;

cos2 θ + sin2 θ = 1

(24/25)2 + sin2 α = 1

sin2α = 1 – (24/25)2

sin2 α = 1 – (576/625) = (625 – 576)/625

sin2 α = (625 – 576)/625 = 49/626

sin α = √49/625 = ±7/25

Hence, sin α = ±7/25.

Problem 2: Prove sin 2A and cos 2A formulae, if ∠A= 30°.

Solution:

Given, ∠A= 30°

We know that,

1) sin 2A = 2 sin A cos A

sin 2(30°) = 2 sin 30° cos 30°

sin 60° = 2 × (1/2) × (√3/2)  {Since, sin 30° = 1/2, cos 30° = √3/2 and sin 60° = √3/2}

√3/2 = √3/2

L.H.S = R.H.S

2) cos 2A = 2cos2A – 1

cos 2(30°) = 2cos2(30°) – 1

cos 60° = 2(√3/2)2 – 1 = 3/2 – 1 {Since, cos 60° = 1/2 and cos 30° = √3/2}

1/2 = 1/2

L.H.S = R.H.S

Hence proved.

Problem 3: Find the value of cos x, if tan x = 3/4.

Solution:

Given, tan x = 3/4

We know that,

tan x = opposite side/adjacent side = 3/4

To find the hypotenuse, we use Pythagoras theorem:

H2= 32 + 42

H2 = 9 + 16 = 25

H = √25 = 5

Now, cos x = adjacent side/hypotenuse

cos x = 4/5

Thus, the value of cos x is 4/5.

Problem 4: Find ∠C (in degrees) and ∠A (in degrees), if ∠B = 45°, BC = 15 in, and AC = 12 in.

Solution:

Given: ∠B = 45°, BC = a = 15 in, and AC = b = 12 in.

From the law of sines, we have

a/sin A = b/sin B = c/sin C

⇒ a/sin A = b/sin B

⇒ 15/sin A = 12/sin 45°

⇒ 15/sin A = 12/(1/√2)

⇒ 15/sin A = 12√2 = 16.97

⇒  sin A = 15/16.97 = 0.8839

⇒  ∠A = sin-1(0.8839) = 62.11°

We know that, sum of interior angles of a triangle is 180°.

So, ∠A + ∠B + ∠C = 180°

⇒ 62.11° + 45° + ∠C = 180°

⇒ ∠C = 180° – (62.11° + 45°) = 72.89°

Hence, ∠A = 62.11° and ∠C = 72.89°.

Problem 5: Prove half-angle identities of the cosine function.

Solution:

The half-angle identity of the cosine function is :

cos (θ/2) = ±√[(1 + cos θ)/2]

From double angle identities, we have,

cos 2A = 2 cos2A – 1

Now replace A with θ/2 on both sides

⇒ cos 2(θ/2) = 2 cos2 (θ/2) – 1

⇒ cos θ = 2 cos2 (θ/2) – 1

⇒ 2cos2(θ/2) = cos θ + 1

⇒ cos2(θ/2) = (cos θ + 1)/2

⇒ cos (θ/2) = ±√[(1 + cos θ)/2]

Hence proved.

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