In any triangle ABC- prove the following-a sinA2sinB - C2 - b sinB2sinC - A2 - c sinC2sinA - B2 - 0

Given: ABC is a triangle, nbsp; [∴ A + B + C = π] According to sine rule asin A= bsin B = csin C = k . . . (i) We have to prove: a sinA2sin(B C2 ...

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Standard XII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

In any triang...

Question

In any triangle ABC, prove the following:

$a sin \frac{A}{2} sin (\frac{B - C}{2}) + b sin \frac{B}{2} sin (\frac{C - A}{2}) + c sin \frac{C}{2} sin (\frac{A - B}{2}) = 0$

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Solution

Given: ABC is a triangle,

$[∴ A + B + C = π]$

According to sine rule

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

We have to prove:

$a sin \frac{A}{2} sin (\frac{B - C}{2}) + b sin \frac{B}{2} sin (\frac{C - A}{2}) + c sin \frac{C}{2} sin (\frac{A - B}{2}) = 0$

$L . H . S = a sin \frac{A}{2} sin (\frac{B - C}{2}) + b sin \frac{B}{2} sin (\frac{C - A}{2}) + c sin \frac{C}{2} sin (\frac{A - B}{2})$

Putting the values from equation (i) in L.H.S

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

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

$= k [sin (B + C) sin \frac{A}{2} sin (\frac{B - C}{2}) + sin (A + C) s i n \frac{B}{2} sin (\frac{C - A}{2}) + sin (A + B) sin \frac{C}{2} sin (\frac{A - B}{2})]$

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

$[∵ sin θ = 2 sin \frac{θ}{2} cos \frac{θ}{2}]$

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

$[∵ A + B + C = π]$

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

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

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

$= 2 k (0)$

= 0

= R.H.S

Hence, proved.

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Trigonometric Ratios Using Right Angled Triangle

Standard XII Mathematics

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In any triangle ABC, prove the following: asinA2sin(B−C2)+bsinB2sin(C−A2)+csinC2sin(A−B2)=0

In any triangle ABC, prove the following:

$a sin \frac{A}{2} sin (\frac{B - C}{2}) + b sin \frac{B}{2} sin (\frac{C - A}{2}) + c sin \frac{C}{2} sin (\frac{A - B}{2}) = 0$