Name: Properties of ITF: Illustration-3
Uploaded: 2019-07-19
Duration: 4 min 28 s
Description: Equations involving inverse tangents can be solved using properties of inverse trigonometric functions. Curious to know what these properties are? Let’s lear...

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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