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

In

△ A B C

, if

a + b + c = 2 s

, then prove that

\sin (\frac{A}{2}) = \sqrt{\frac{(s - b) (s - c)}{b c}}

, with usual notations.

Accepted Answer

Given: $a + b + c = 2 s$

To prove: $\sin (\frac{A}{2}) = \sqrt{\frac{(s - b) (s - c)}{b c}}$

We know that $△ A B C = \frac{1}{2} \cdot b c \cdot \sin A . . . (1)$ and also using herons formula $Δ A B C = \sqrt{s (s - a) (s - b) (s - c)} . . . (2)$

Using $(1)$ and $(2)$ , we get:

$\frac{1}{2} \cdot b c \cdot \sin A = \sqrt{s (s - a) (s - b) (s - c)}$

$\Rightarrow \sin A = \frac{2 \sqrt{s (s - a) (s - b) (s - c)}}{b c}$

$\Rightarrow 2 \sin \frac{A}{2} \cos \frac{A}{2} = \frac{2 \sqrt{s (s - a) (s - b) (s - c)}}{b c} [∵ \sin 2 θ = 2 \sin θ \cos θ]$

$\Rightarrow \sin \frac{A}{2} \cos \frac{A}{2} = \frac{\sqrt{s (s - a) (s - b) (s - c)}}{b c} . . . (*)$

Now, we know that $\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ [using cosine rule] and $\cos 2 A = 2 \cos^{2} A - 1$ .

$\Rightarrow 2 \cos^{2} \frac{A}{2} - 1 = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$

$\Rightarrow 2 \cos^{2} \frac{A}{2} = \frac{b^{2} + c^{2} - a^{2}}{2 b c} + 1$

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

$\Rightarrow \cos^{2} \frac{A}{2} = \frac{b^{2} + c^{2} - a^{2} + 2 b c}{4 b c}$

$\Rightarrow \cos^{2} \frac{A}{2} = \frac{(b + c - a) (b + c + a)}{4 b c}$

$\Rightarrow \cos^{2} \frac{A}{2} = \frac{(2 s - a - a) (2 s)}{4 b c} [∵ b + c + a = 2 s]$

$\Rightarrow \cos^{2} \frac{A}{2} = \frac{(s - a) (4 s)}{4 b c}$

$\Rightarrow \cos^{2} \frac{A}{2} = \frac{s (s - a)}{b c}$

$\Rightarrow \cos \frac{A}{2} = \sqrt{\frac{s (s - a)}{b c}} . . . (3)$

Putting the equation $(3)$ in equation $(*)$ , we get:

$\sin \frac{A}{2} \sqrt{\frac{s (s - a)}{b c}} = \frac{\sqrt{s (s - a) (s - b) (s - c)}}{b c}$

$\Rightarrow \sin \frac{A}{2} = \sqrt{\frac{(s - b) (s - c)}{b c}}$

Hence proved.

In $△ A B C$ , if $a + b + c = 2 s$ , then prove that $\sin (\frac{A}{2}) = \sqrt{\frac{(s - b) (s - c)}{b c}}$ , with usual notations.

Important Questions on Trigonometric Functions

Learn and Prepare for any exam you want !

In △ABC, if a+b+c=2s, then prove that sin⁡A2=(s−b)(s−c)bc, with usual notations.

Important Questions on Trigonometric Functions

Learn and Prepare for any exam you want !

In $△ A B C$ , if $a + b + c = 2 s$ , then prove that $\sin (\frac{A}{2}) = \sqrt{\frac{(s - b) (s - c)}{b c}}$ , with usual notations.