What are Cosine Formulas?

Trigonometry is a discipline of mathematics that studies the relationships between the lengths of the sides and angles of a right-angled triangle. Trigonometric functions, also known as goniometric functions, angle functions, or circular functions, are functions that establish the relationship between an angle to the ratio of two of the sides of a right-angled triangle. The six main trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant. Trigonometric angles are the Angles defined by the ratios of trigonometric functions. Trigonometric angles represent trigonometric functions. The value of the angle can be anywhere between 0-360°.

👁 Right Angle Triangle

As given in the above figure in a right-angled triangle:

• Hypotenuse: The side opposite to the right angle is the hypotenuse, It is the longest side in a right-angled triangle and opposite to the 90° angle.
• Base: The side on which angle C lies is known as the base.
• Perpendicular: It is the side opposite to angle C in consideration.

Trigonometric Ratios Definition

Trigonometry has 6 basic trigonometric ratios also called trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows,

• sine: the ratio of perpendicular and hypotenuse is defined as sine and It is represented as sin θ
• cosine: the ratio of base and hypotenuse is defined as cosine and it is represented as cos θ
• tangent: the ratio of sine and cosine of an angle is defined as a tangent. Thus the definition of tangent comes out to be the ratio of perpendicular and base and is represented as tan θ.
• cosecant: It is the reciprocal of sin θ and is represented as cosec θ.
• secant: It is the reciprocal of cos θ and is represented as sec θ.
• cotangent: It is the reciprocal of tan θ and is represented as cot θ.

According to the above image, Trigonometric Ratios are

• sin θ = Perpendicular/Hypotenuse = AB/AC
• cos θ = Base/Hypotenuse = BC/AC
• tan θ = Perpendicular/Base = AB/BC
• cosec θ = Hypotenuse/Perpendicular = AC/AB
• sec θ = Hypotenuse/Base = AC/BC
• cot θ = Base/Perpendicular = BC/AB

Formulas and Identities for Cosine

There are various different identities in trigonometry, some of these identities for cosine trigonometric ratio are discussed as follows:

Reciprocal Identities

• sin θ = 1/ cosec θ
• cosec θ = 1/ sin θ
   
• cos θ = 1/ sec θ 
• sec θ = 1 / cos θ
   
• cot θ = 1 / tan θ
• tan θ = 1 / cot θ
   
• cot θ = cos θ / sin θ
• tan θ = sin θ / cos θ
   
• tan θ.cot θ = 1

Complementary Angles Identities 

Pair of angles whose sum is equal to 90° are called complementary angles and the identities of Complementary angles are:

• sin (90° - θ) = cos θ
• cos (90° - θ) = sin θ
• tan (90° - θ) = cot θ
• cot (90° - θ) = tan θ
• sec (90° - θ) = cosec θ
• cosec (90° - θ) = sec θ

Supplementary Angles Identities

Pair of angles whose sum is equal to 180° are called supplementary angles and the identities of supplementary angles are:

• sin (180° - θ) = sin θ
• cos (180° - θ) = - cos θ
• tan (180° - θ) = - tan θ
• cot (180° - θ) = - cot θ
• sec (180° - θ) = - sec θ
• cosec (180° - θ) = - cosec θ

Cosine Formulas Using Pythagorean Identity

One of the trigonometric identities between sin and cos. It represents  sin²x + cos²x = 1

 sin²x + cos²x = 1

Now Subtracting sin²x from both sides,

cos²x = 1 - sin²x

now square both sides

cos x = ± √(1 - sin²x)

Cosine Formulas with Sum/Difference Formulas

There are sum/difference formulas for every trigonometric function that deal with the sum of angles (x + y) and the difference of angles (x - y). 

Formulas of cosine function with sum difference formulaes are,

cos(x + y) = cos (x) cos(y) – sin (x) sin (y)

cos (x – y) = cos (x) cos (y) + sin (x) sin (y)

Formulas for Law of Cosines

This law is used to find the missing sides/angles in a non-right angled triangle. Assume a triangle ABC in which AB = c, BC = a, and CA = b. 

The cosine formulas  are,

cos A = (b² + c² - a²)/(2bc)

cos B = (c² + a² - b²)/(2ac)

cos C = (a² + b² - c²)/(2ab)

Learn more about, Law of Cosine Formula

Double Angle Formula of Cosine

 In trigonometry while dealing with 2 times the angle. There are multiple sorts of double-angle formulas of cosine and from that, we use one of the following while solving the problem depending on the available information. 

cos 2x = cos²(x) – sin²(x)

cos 2x = 2 cos²(x) − 1  

cos 2x = 1 – 2 sin²(x)

cos 2x = [(1 - tan²x)/(1 + tan²x)]

Half Angle Formula of Cosine

Half angle formula for the cosine is similar to the double angle formula but 2(Angle) is changed to (Angle). The half-angle formula for the cosie is,

cos (x/2) =± √[ (1 + cos x) / 2]

Triple Angle Formula of Cosine

The triple angle formula for the cosine is,

cos 3x = 4cos³x - 3cosx 

Read More,

• Trigonometric Identities
• Trigonometric Table

Sample Problems on Cosine Formulas

Problem 1: If sin a = 3/5 and a is in the first quadrant, find the value of cos a.

Solution:

Using one of the cosine formulas,

cos a = ± √(1 - sin²a)

Since a is in the first quadrant, cos a is positive. Thus,

cos a = √(1 - sin²a)

Substitute sin a = 3/5 here,

cos a = √(1 - (3/5)²)  

⇒ cos a = √(1 - 9/25)

⇒ cos a =√ (16/25)

Thus, cos a = 4/5

Problem 2: If sin (90 - A) = 2/3, then find the value of cos A.

Solution:

Using one of the cosine formulas,

cos A = sin (90 - A)

given that sin (90 - A) = 2/3. Hence,

cos A = 2/3

The value of cos A is 2/3.

Problem 3: In a triangle ABC, AB = c, BC = a, and CA = b. Also, a = 50 units, b = 60 units, and c = 30 units. Find cos A.

Solution:

By Using the cosine formula of law of cosines,

cos A = (b² + c² - a²) / (2bc)

⇒ cos A = (60² + 30² - 50²) / (2 · 60 · 30)

⇒ cos A = (3600 + 900 - 2500) / 3600

⇒ cos A = 2000 / 3600

Thus, cos A = 5/9

Problem 4: If cos A = 4/5, cos B = 12/13, find the value of cos (A+B)?

Solution: 

Here given cos A = 4/5, cos B = 12/13

since A and B both lie in 4th quadrant and in 4th quadrant Sin A and Sin B will be negative.

therefore, 

sin A  = - √(1 - cos² A)

⇒ sin A =  √{1 - (4/5)² }

⇒ sin A =  - √(1 - 16/25)

⇒ sin A =  -3/5

and sin B = - √(1 - cos² B)

⇒ sin B =  - √{1 - (12/13)²}

⇒ sin B =  -5/13

Now, as per the formula 

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

⇒ cos(A + B) = 4/5 × 12/13 - (-3/5)(-5/13)

⇒ cos(A + B) = 48/65 - 15/65

⇒ cos(A + B) = 33/65

Problem 5: Prove that cos4x = 1- 8sin²xcos²x.

Solution: 

Given that 

LHS = cos4x

⇒ LHS = cos2(2x) 

⇒ LHS = cos 2x     [As cos 2x = 1 – 2 sin²(x)]

⇒ LHS = 1 – 2 sin²  2(x)

⇒ LHS = 1 - 2 (sin2x)²

⇒ LHS = 1 - 2(2sinx cosx)²

⇒ LHS =  1 -  8sin²xcos²x 

⇒ LHS =  RHS 

[Hence Proved]