Answer on Question #44669 – Math - Trigonometry
Prove that in a triangle with angles A, B and C; and sides of length a, b and c that:
1 / [ ( a − b ) ( a − c ) ] ∗ tan ( A / 2 ) + 1 / [ ( b − c ) ( b − a ) ] ∗ tan ( B / 2 ) + 1 / [ ( c − b ) ( c − a ) ] ∗ tan ( C / 2 ) = (Area of triangle) ( − 1 ) 1 / [(a - b)(a - c)]^* \tan(A/2) + 1 / [(b - c)(b - a)]^* \tan(B/2) + 1 / [(c - b)(c - a)]^* \tan(C/2) = \text{(Area of triangle)}^(-1) 1/ [( a − b ) ( a − c ) ] ∗ tan ( A /2 ) + 1/ [( b − c ) ( b − a ) ] ∗ tan ( B /2 ) + 1/ [( c − b ) ( c − a ) ] ∗ tan ( C /2 ) = (Area of triangle) ( − 1 )
{NOTE: 'a' is the length of side opposite to angle A, likewise 'b' is the length of side opposite to angle B and similarly 'c' is the length of side opposite to angle C.}
Solution.
According to the Law of Cosines:
cos A = b 2 + c 2 − a 2 2 b c \cos A = \frac{b^2 + c^2 - a^2}{2bc} cos A = 2 b c b 2 + c 2 − a 2
As we know
cos A = 2 cos 2 A 2 − 1 → cos 2 A 2 = 1 + cos A 2 = b 2 + c 2 − a 2 + 2 b c 4 b c = ( b + c − a ) ( b + c + a ) 4 b c = s ( s − a ) b c , where \cos A = 2 \cos^2 \frac{A}{2} - 1 \rightarrow \cos^2 \frac{A}{2} = \frac{1 + \cos A}{2} = \frac{b^2 + c^2 - a^2 + 2bc}{4bc} = \frac{(b + c - a)(b + c + a)}{4bc} = \frac{s(s - a)}{bc}, \quad \text{where} cos A = 2 cos 2 2 A − 1 → cos 2 2 A = 2 1 + cos A = 4 b c b 2 + c 2 − a 2 + 2 b c = 4 b c ( b + c − a ) ( b + c + a ) = b c s ( s − a ) , where s = a + b + c 2 . s = \frac{a + b + c}{2}. s = 2 a + b + c . cos A 2 = s ( s − a ) b c , sin A 2 = ( s − b ) ( s − c ) b c , \cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}, \quad \sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}}, cos 2 A = b c s ( s − a ) , sin 2 A = b c ( s − b ) ( s − c ) , tan A 2 = ( s − b ) ( s − c ) s ( s − a ) = P s ( s − a ) \tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} = \frac{P}{s(s - a)} tan 2 A = s ( s − a ) ( s − b ) ( s − c ) = s ( s − a ) P
Where P = s ( s − a ) ( s − b ) ( s − c ) P = \sqrt{s(s - a)(s - b)(s - c)} P = s ( s − a ) ( s − b ) ( s − c )
Similarly: tan B 2 = P s ( s − b ) \tan \frac{B}{2} = \frac{P}{s(s - b)} tan 2 B = s ( s − b ) P tan C 2 = P s ( s − c ) \tan \frac{C}{2} = \frac{P}{s(s - c)} tan 2 C = s ( s − c ) P
So, tan 2 A ( a − b ) ( a − c ) + tan 2 B ( b − c ) ( b − a ) + tan c C ( c − b ) ( c − a ) = \frac{\tan_2^A}{(a - b)(a - c)} + \frac{\tan_2^B}{(b - c)(b - a)} + \frac{\tan_c^C}{(c - b)(c - a)} = ( a − b ) ( a − c ) t a n 2 A + ( b − c ) ( b − a ) t a n 2 B + ( c − b ) ( c − a ) t a n c C =
= P s [ 1 ( a − b ) ( a − c ) ( s − a ) + 1 ( b − a ) ( b − c ) ( s − b ) + 1 ( c − a ) ( c − b ) ( s − c ) ] = = \frac{P}{s} \left[ \frac{1}{(a - b)(a - c)(s - a)} + \frac{1}{(b - a)(b - c)(s - b)} + \frac{1}{(c - a)(c - b)(s - c)} \right] = = s P [ ( a − b ) ( a − c ) ( s − a ) 1 + ( b − a ) ( b − c ) ( s − b ) 1 + ( c − a ) ( c − b ) ( s − c ) 1 ] = = P s ∗ − ( b − c ) ( s − b ) ( s − c ) − ( c − a ) ( s − a ) ( s − c ) − ( a − b ) ( s − a ) ( s − b ) ( a − b ) ( b − c ) ( c − a ) ( s − a ) ( s − b ) ( s − c ) = = − P s ( s − a ) ( s − b ) ( s − c ) ∗ ∗ ( b − c ) ( a − b + c ) ( a + b − c ) + ( c − a ) ( b + c − a ) ( a + b − c ) + ( a − b ) ( b + c − a ) ( a − b + c ) 4 ( a − b ) ( b − c ) ( c − a ) = − P P 2 ∗ 4 a c 2 + 4 a 2 b − 4 a 2 c + 4 b 2 c − 4 b c 2 + 4 a b 2 4 ( a − b ) ( b − c ) ( c − a ) = 1 P ∗ 4 ( a − b ) ( b − c ) ( c − a ) 4 ( a − b ) ( b − c ) ( c − a ) = = 1 P . \begin{array}{l}
= \frac {P}{s} * \frac {- (b - c) (s - b) (s - c) - (c - a) (s - a) (s - c) - (a - b) (s - a) (s - b)}{(a - b) (b - c) (c - a) (s - a) (s - b) (s - c)} = \\
= - \frac {P}{s (s - a) (s - b) (s - c)} * \\
* \frac {(b - c) (a - b + c) (a + b - c) + (c - a) (b + c - a) (a + b - c) + (a - b) (b + c - a) (a - b + c)}{4 (a - b) (b - c) (c - a)} \\
= - \frac {P}{P ^ {2}} * \frac {4 a c ^ {2} + 4 a ^ {2} b - 4 a ^ {2} c + 4 b ^ {2} c - 4 b c ^ {2} + 4 a b ^ {2}}{4 (a - b) (b - c) (c - a)} = \frac {1}{P} * \frac {4 (a - b) (b - c) (c - a)}{4 (a - b) (b - c) (c - a)} = \\
= \frac {1}{P}.
\end{array} = s P ∗ ( a − b ) ( b − c ) ( c − a ) ( s − a ) ( s − b ) ( s − c ) − ( b − c ) ( s − b ) ( s − c ) − ( c − a ) ( s − a ) ( s − c ) − ( a − b ) ( s − a ) ( s − b ) = = − s ( s − a ) ( s − b ) ( s − c ) P ∗ ∗ 4 ( a − b ) ( b − c ) ( c − a ) ( b − c ) ( a − b + c ) ( a + b − c ) + ( c − a ) ( b + c − a ) ( a + b − c ) + ( a − b ) ( b + c − a ) ( a − b + c ) = − P 2 P ∗ 4 ( a − b ) ( b − c ) ( c − a ) 4 a c 2 + 4 a 2 b − 4 a 2 c + 4 b 2 c − 4 b c 2 + 4 a b 2 = P 1 ∗ 4 ( a − b ) ( b − c ) ( c − a ) 4 ( a − b ) ( b − c ) ( c − a ) = = P 1 .
Q.E.D.
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