Cosine rules

The cosine rule is frequently used in trigonometry. It is commonly known as the law of cosines or simply, cosine formula. The cosine rule states that the sum of the squares of the length of other sides and twice the product of other sides with their cosine angles, rather than hypotenuse, is equal to the third side of the triangle.

Let us take a triangle ABC with angles x, y, and z, respectively. According to the cosine rule, we get the following formulae-

• a² = b² + c² – 2bc cos ∠x
• b² = a² + c² – 2ac cos ∠y
• c² = a² + b² – 2ab cos ∠z

These formulae can be written as-

• cos x = (b² + c² -a²)/2bc
• cos y = (a² + c² -b²)/2ac
• cos z = (a² + b² – c²)/2ab

Cosine rule is helpful to get the length of the side of the triangle by knowing their angles. We can directly find the angles and sides of a triangle from this rule.

Proof of cosine rule

Let us consider a triangle with sides a, b and c respective to their corners A, B and C. We need to draw a perpendicular from B on the side AC at D, as shown in the figure above.

According to the trigonometry ratio and from triangle BCD, we get,

cos C = CD/a [cos θ = Base/Hypotenuse]

This can be re-written as-

CD = a cos C ………… (1)

Subtracting equation 1 from side b on both the sides, we get;

b – CD = b – a cos C

or it can be written as,

DA = b – a cos C

Again, according to the trigonometry ratio and from triangle BCD, we get,

sin C = BD/a [sin θ = Perpendicular/Hypotenuse]

This can be re-written as,

BD = a sin C ……….(2)

By using Pythagoras theorem in triangle ADB, we get;

c² = BD² + DA² [Hypotenuse² = Perpendicular² + Base²]

Substituting the value of DA and BD from equation 1 and 2, we get;

c² = (a sin C)² + (b – a cos C)²

c² = a² sin²C + b² – 2ab cos C + a² cos² C

c² = a² (sin²C + cos² C) + b² – 2ab cos C

By trigonometric identities, we know;

sin²θ+ cos²θ = 1

Therefore,

c² = a² + b² – 2ab cos C

Hence, proved.

Derivation of cosine formula from law of sines

Let us consider a triangle with side a, b, c and their respective angles by x, y and z.

We know, from the law of sines,

c/sin z = b/sin y = a/sin x

Also, the sum of angles inside a triangle is equal to 180 degrees, i.e. equal to π.

Therefore, x+y+z = π

Using the third equation system, we get

c/sin z = b/sin (x+z) ----------- (1)

c/sin z = a/sin x

Using angle sum and difference identities, we get,

sin (x+z) = sin x cos z + sin z cos x

c (sin x cos z + sin z cos x) = b sin z

c sin x = a sin z

Dividing the whole equation by cos z, we get,

c (sin x + tan z cos x) = b tan z

c sin x/cos z = a tan z

c² sin²x / cos²z = tan z

From equation 1, we get,

c sin x / b – c cos x = tan z

1 + tan²z = 1/cos²z

c² sin²x (1+ (c² sin² x / (b – c cos x)²)) = a² (c² sin² x / (b – c cos x)²)

Multiplying the equation by (b – c cos x)² and rearranging it, we get,

a² = b² + c² – 2bc cos x.

Hence, proved.