Question

sinA +sinB - sinC =4sinA/2 sinB/2 cosC/2

Answer 1

this is the sol. than u for patience

Answer 2

Answer:

$sinA+sinB-sinC=4sin\frac{A}{2} sin\frac{B}{2} cos\frac{C}{2}$

Step-by-step explanation:

$LHS = sinA+sinB-sinC$

$= 2sin\left(\frac{A+B}{2}\right) cos\left(\frac{A-B}{2}\right)- 2sin\frac{C}{2} cos\frac{C}{2}$

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We know that,

$i)sinA+sinB = 2sin\left(\frac{A+B}{2}\right) sin\left(\frac{A-B}{2}\right)\\ii)sinC = 2sin\frac{C}{2} cos\frac{C}{2}$

$iii) A+B+C = 180$

$\implies A+B = 180-C\\\implies \frac{A+B}{2}=90-\frac{C}{2}\\\implies sin\left(\frac{A+B}{2}\\=sin\left(90-\frac{C}{2}\right)\\=cos\frac{C}{2}$

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$=2cos\frac{C}{2} cos\left(\frac{A-B}{2}\right)- 2sin\frac{C}{2} cos\frac{C}{2}$

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

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

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

$=2cos\frac{C}{2} [2sin\frac{2A}{4} sin\frac{(A+B-A+B)}{4} ]$

$=2cos\frac{C}{2} [2sin\frac{2A}{4} sin\frac{2B}{4} ]$

$=2cos\frac{C}{2} [2sin\frac{A}{2} sin\frac{B}{2} ]$

$=4cos\frac{C}{2} sin\frac{A}{2} sin\frac{B}{2}$

$=4sin\frac{A}{2} sin\frac{B}{2} cos\frac{C}{2}$

Therefore,

$sinA+sinB-sinC=4sin\frac{A}{2} sin\frac{B}{2} cos\frac{C}{2}$

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