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Mathematics

The Cosine Formulae

Gk Scientist March 11, 2022 No Comments
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The Cosine Formulae:

In △ABC with BC = a, CA = b and AB = c,

(i) cos A = (b2 + c2 – a2)/2bc (ii) cos B = (c2 + a2 – b2)/2ca (iii) cos C = (a2 + b2 – c2)/2ab

cosine formulae in trigonometry - The Cosine Formulae

Proof- The proofs of all three formulae are similar. Let us take the first one. We consider three different types of △ABC- (i) acute-angled (ii) right-angled at A and (iii) obtuse-angled at A.

Case (i):

CD is drawn perpendicular to AB, In △CAD, CD/AC = sin A [Fig. (i)]
∴ CD = AC sin A = b sin A
Now, DB = c – AD = c – (AD/AC)AC = c – b cos A

Now, in right-angled △CDB,
BC2 = CD2 + BD2
⇒ a2 = (b sin A)2 + (c – b cos A)2
⇒ a2 = b2 sin2 A + c2 – 2bc cos A + b2 cos2 A
⇒ a2 = b2 (sin2 A + cos2 A) + c2 – 2 bc cos A
⇒ a2 = b2 + c2 – 2 bc cos A

∴ cos A = (b2 + c2 – a2)/2bc

Case (ii):

In △ABC, right-angled at A [Fig. (ii)]
a2 = b2 + c2 = b2 + c2 – 2bc cos A (∵ cos A = cos 90° = 0)
∴ 2 bc cos A = b2 + c2 – a2

∴ cos A = (b2 + c2 – a2)/2bc

Case (iii):

In △ABC, ∠A > 90°. CD is drawn perpendicular to BA produced [Fig. (iii)]

∴ In right-angled triangle CDA,
CD/AC = sin A, ∴ CD = AC sin A = b sin A
Also, AD/AC = cos A, ∴ AD = AC cos A = b cos A

Now, in right-angled triangle CDB,
BC2 = CD2 + DB2
∴ a2 = (b sin A)2 + (b cos A + c)2
⇒ a2 = b2 sin2 A + b2 cos2 A + 2bc cos A + c2
⇒ a2 = b2 + c2 + 2bc cos A

∴ 2bc cos A = b2 + c2 – a2
∴ cos A = (b2 + c2 – a2)/2bc

From the above three cosine formulae, one can write-

a2 = b2 + c2 – 2bc cos A
b2 = c2 + a2 – 2ca cos B
c2 = a2 + b2 – 2ab cos C

Adding up, a2 + b2 + c2 = 2 (a2 + b2 + c2) – 2 (bc cos A + ca cos B + ab cos C)

⇒ a2 + b2 + c2 = 2 (bc cos A + ca cos B + ab cos C)

As a2 + b2 + c2 > 0, bc cos A + ca cos B + ab cos C > 0


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