Name: The Projection Formula
Uploaded: 2019-07-19
Duration: 3 min 44 s
Description: The projection formula involves the scalar projection and vector projection. In this video, we will learn about the projection formula and solve problems bas...

Question

Using the Sine rule, prove the Cosine rule.

Accepted Answer

Let $A B C$ be a triangle whose sides $A B, B C,$ and $A C$ are of length $c$ units, $a$ units, and $b$ units.

According to the sine rule we have

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k . . . . . (i)$

$\Rightarrow a = k \sin A, b = k \sin B,$ and $c = k \sin C$

Now, according to the cosine rule, we have

$\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$

i.e., we need to prove $a^{2} + b^{2} - c^{2} = 2 a b \cos C$

Consider

L.H.S. $= a^{2} + b^{2} - c^{2}$

$= {(k \sin A)}^{2} + {(k \sin B)}^{2} - {(k \sin C)}^{2}$

$= k^{2} (\sin^{2} A + \sin^{2} B - \sin^{2} C)$

$= \frac{k^{2}}{2} (2 \sin^{2} A + 2 \sin^{2} B - 2 \sin^{2} C)$

$= \frac{k^{2}}{2} (1 - \cos 2 A + 2 \sin^{2} B - 1 + \cos 2 C)$ $[∵ \cos 2 A = 1 - 2 \sin^{2} A]$

$= \frac{k^{2}}{2} (\cos 2 C - \cos 2 A + 2 \sin^{2} B)$

$= \frac{k^{2}}{2} (2 \sin (\frac{2 C + 2 A}{2}) \sin (\frac{2 A - 2 C}{2}) + 2 \sin^{2} B)$ $[∵ \cos C - \cos D = 2 \sin (\frac{C + D}{2}) \sin (\frac{D - C}{2})]$

$= \frac{k^{2}}{2} (2 \sin (C + A) \sin (A - C) + 2 \sin^{2} B)$

$= k^{2} (\sin (C + A) \sin (A - C) + \sin^{2} B)$

$= k^{2} (\sin (π - B) \sin (A - C) + \sin^{2} B)$ $[∵ A + B + C = π]$

$= k^{2} (\sin B \sin (A - C) + \sin^{2} B)$

$= k^{2} \sin B (\sin (A - C) + \sin B)$

$= k^{2} \sin B (\sin (A - C) + \sin (A + C))$

$= k^{2} \sin B (2 \sin (\frac{2 A}{2}) \cos (\frac{- 2 C}{2}))$ $[∵ \sin C + \sin D = 2 \sin (\frac{C + D}{2}) \cos (\frac{C - D}{2})]$

$= k^{2} \sin B (2 \sin (A) \cos (C))$

$= 2 k \sin B \cdot k \sin (A) \cos (C)$

$= 2 a b \cos C$ [Since from $(i)$ , we have $k \sin A = a,$ and $k \sin B = b$ ]

Using the Sine rule, prove the Cosine rule.

Important Questions on Trigonometric Functions

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