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If in a Triangle ABC b , c , B ( where A ,B ,C denotes angles and a,b,c denotes sides of the triangle ABC ) are given and b < c prove that $$ \ sin \frac {(A_1-A_2)} {2}\ = \frac { (a_1-a_2)} {(2b)} $$

MY ATTEMPT :

taking cosine rule $$ cos B = \frac{\ a^2 + c^2 - b^2} {2ac} $$ and rearranging the terms we get $${a^2 -2ac cosB +(c^2-b^2)} = 0 $$ taking $ a_1 $ and $a_2$ as roots and solving $ a_1 $ + $a_2$ = $ 2c$ $cos B$ and $a_1$ $a_2$ = $ c^2$ - $b^2$ we get $$ ( a_1 - a_2 )^2 = 4b^2 -4c^2 cos^2 B $$

I am struck here please help me to solve this question

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    $\begingroup$ It is not clear what $A_1$, $A_2$ etc. is. Can you clarify by using $A$, $B$, $C$ for angles and $a$, $b$, $c$ for corresponding facing sides? $\endgroup$
    – mf67
    Dec 31, 2020 at 17:01
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    $\begingroup$ @mf67 I think it means the two possible length of $A$ as $b,c,B$ does not define one triangle but two possible triangles . $\endgroup$
    – cr001
    Dec 31, 2020 at 17:07
  • $\begingroup$ I do not think it is correct unless I have understood the question wrong. What is true is $\ sin {\frac{A_1-A_2} {2}} = \frac { (a_1-a_2)} {(2b)}$ $\endgroup$
    – Math Lover
    Dec 31, 2020 at 17:24
  • $\begingroup$ Actually I think the result is wrong. The result only holds when $C=30^{\circ}$ i.stack.imgur.com/YYF8y.png $\endgroup$
    – cr001
    Dec 31, 2020 at 17:25
  • $\begingroup$ Given $\angle B$ is fixed and so are sides $b, c$, you get two values of $A$ when the side opposite $B$ is equal to $b$. It is possible because $b \lt c$. There will be values of $b, c, B$ that will give you a right angled triangle and then $A_1 - A_2 = 0$ but that is a special case. $\endgroup$
    – Math Lover
    Dec 31, 2020 at 17:28

1 Answer 1

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enter image description here

As shown in the picture, $BC=a_1, BC'=a_2, \theta = {A_1 - A_2\over 2}$.

Therefore $\sin({A_1 - A_2\over 2}) = {{a_1-a_2\over 2}\over b}$ by definition of sine function in right triangle.

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  • $\begingroup$ thanks but cant we solve it manually I mean by using cosine rule etc $\endgroup$
    – victor
    Dec 31, 2020 at 17:42
  • $\begingroup$ @victor ${\sin(A_1)\over \sin(B)}= {a_1\over b}$ and ${\sin(A_2)\over \sin(B)}= {a_2\over b}$ so ${\sin(A_1)-\sin(A_2)\over 2\sin(B)}={a_1-a_2\over 2b}$. You can then use the formula $\sin(A_1)-\sin(A_2)=2\sin({A_1-A_2\over 2})\cos({A_1+A_2\over 2})$ and show $\cos({A_1+A_2\over 2})=\sin(B)$. But this is way too complicated compared to the geometric method. $\endgroup$
    – cr001
    Dec 31, 2020 at 17:55

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