# 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°.

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 Functions**

Trigonometry has 6 basic 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

cosine θ = Base/Hypotenuse = BC/ACtangent θ = Perpendicular/Base = AB/BC

cosecant θ = Hypotenuse/Perpendicular = AC/AB

secant θ = Hypotenuse/Base = AC/BC

cotangent θ = Base/Perpendicular = BC/AB

**Reciprocal Identities**

sin θ = 1/ cosec θ OR cosec θ = 1/ sin θ

cos θ = 1/ sec θ OR sec θ = 1 / cos θcot θ = 1 / tan θ OR tan θ = 1 / cot θ

cot θ = Cos θ / sin θ OR tan θ = sin θ / cos θ

tan θ.cot θ = 1

**Trigonometric Identities of Complementary and Supplementary Angles**

**Complementary Angles:** Pair of angles whose sum is equal to 90°. Identities of Complementary angles are:

sin (90° – θ) = cos θ

cos (90° – θ) = sin θtan (90° – θ) = cot θ

cot (90° – θ) = tan θ

sec (90° – θ) = cosec θ

cosec (90° – θ) = sec θ

**Supplementary Angles:** Pair of angles whose sum is equal to 180°. Identities of supplementary angles are:

sin (180° – θ) = sin θ

cos (180° – θ) = – cos θtan (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 sin

^{2}x + cos^{2}x = 1

sin^{2}x + cos^{2}x = 1Now Subtracting sin

^{2}x from both sides,cos

^{2}x = 1 – sin^{2}xnow square both sides

cos x = ± √(1 – sin^{2}x)

**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 of Cosine Using 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 = (b^{2}+ c^{2 }– a^{2})/(2bc)

cos B = (c^{2}+ a^{2}– b^{2})/(2ac)

cos C = (a^{2}+ b^{2}– c^{2})/(2ab)

**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 = cos^{2}(x) – sin^{2}(x)

cos 2x = 2 cos^{2}(x) − 1

cos 2x = 1 – 2 sin^{2}(x)

cos 2x = [(1 – tan^{2}x)/(1 + tan^{2}x)]

**Triple Angle Formula of cosine**

cos 3x = 4cos^{3}x – 3cosx

### Sample Problems

**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 – sin^{2}a)Since a is in the first quadrant, cos a is positive. Thus,

cos a = √(1 – sin

^{2}a)Substitute sin a = 3/5 here,

cos a = √(1 – (3/5)

^{2})cos a = √(1 – 9/25)

cos a =√ (16/25)

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 1/2.

**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 = (b^{2}+ c^{2}– a^{2}) / (2bc)= (60

^{2}+ 30^{2}– 50^{2}) / (2 · 60 · 30)= (3600 + 900 – 2500) / 3600

= 2000 / 3600

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 – cos

^{2}A)= √{1 – (4/5)

^{2}}= – √(1 – 16/25)

= -3/5

Sin B = – √(1 – cos

^{2}B)= – √{1 – (12/13)

^{2}}= -5/13

now

As per the formulae

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

= 48/65 – 15/65

= 33/65

**Problem 5: Prove that cos4x = 1- 8sin ^{2}xcos^{2}x.**

**Solution: **

Given that

LHS = cos4x

= cos2(2x)

= cos 2x

{cos 2x = 1 – 2 sin^{2}(x)}= 1 – 2 sin

^{2 }2(x)= 1 – 2 (sin2x)

^{2}

^{ }= 1 – 2(2sinx cosx)^{2}

^{ }= 1 – 8sin^{2}xcos^{2}x= RHS

Hence proved

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