Question

A+B+C=0° Sina+SinB +sinc=ksina/2 sinb/2sinc/2

then k is.
​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\mapsto\pink{A+B+C=0^{\circ}}}$

Now,

$\sf{sin(A)+sin(B)+sin(C)}$

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

$\sf{=-sin(B+C)+2\,sin\left(\dfrac{B+C}{2}\right)\,cos\left(\dfrac{B-C}{2}\right)}$

$\sf{=2\,sin\left(\dfrac{B+C}{2}\right)\,cos\left(\dfrac{B-C}{2}\right)-sin(B+C)}$

$\sf{=2\,sin\left(\dfrac{B+C}{2}\right)\,cos\left(\dfrac{B-C}{2}\right)-2\,sin\left(\dfrac{B+C}{2}\right)\,cos\left(\dfrac{B+C}{2}\right)}$

$\sf{=2\,sin\left(\dfrac{B+C}{2}\right)\left\{cos\left(\dfrac{B-C}{2}\right)-cos\left(\dfrac{B+C}{2}\right)\right\}}$

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

$\sf{=-2\,sin\left(\dfrac{A}{2}\right)\left\{2\,sin\left(\dfrac{B}{2}\right)\,sin\left(\dfrac{C}{2}\right)\right\}}$

$\sf{=-4\,sin\left(\dfrac{A}{2}\right)\,sin\left(\dfrac{B}{2}\right)\,sin\left(\dfrac{C}{2}\right)}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\red{\rm :\longmapsto\:A + B + C = 0}$

and

$\red{\rm :\longmapsto\:sinA + sinB + sinC = ksin\bigg[\dfrac{A}{2} \bigg]sin\bigg[\dfrac{B}{2} \bigg]sin\bigg[\dfrac{C}{2} \bigg]}$

Let us consider,

$\rm :\longmapsto\:sinA + sinB + sinC$

$\rm \:  =  \: (sinA + sinB) + sinC$

$\rm \:  =  \: 2sin\bigg[\dfrac{A + B}{2} \bigg]cos\bigg[\dfrac{A - B}{2} \bigg] + sinC$

As it is given that,

$\red{\rm :\longmapsto\:A + B + C = 0\rm \implies\:A + B = - C}$

Also, we know that

$\red{\rm :\longmapsto\:sin2x = 2sinx \: cosx \: }$

So, using this identity, we get

$\rm \:  =  \: 2sin\bigg[\dfrac{ - C}{2} \bigg]cos\bigg[\dfrac{A - B}{2} \bigg] + 2sin\bigg[\dfrac{C}{2} \bigg]cos\bigg[\dfrac{C}{2} \bigg]$

$\rm \:  =  \: - 2sin\bigg[\dfrac{C}{2} \bigg]cos\bigg[\dfrac{A - B}{2} \bigg] + 2sin\bigg[\dfrac{C}{2} \bigg]cos\bigg[\dfrac{C}{2} \bigg]$

$\rm \:  =  \: 2sin\bigg[\dfrac{C}{2} \bigg]\bigg( - cos\bigg[\dfrac{A - B}{2} \bigg] + cos\bigg[\dfrac{C}{2} \bigg]\bigg)$

$\rm \:  =  \: 2sin\bigg[\dfrac{C}{2} \bigg]\bigg( - cos\bigg[\dfrac{A - B}{2} \bigg] + cos\bigg[\dfrac{ - (A + B)}{2} \bigg]\bigg)$

We know

$\rm \:  =  \: 2sin\bigg[\dfrac{C}{2} \bigg]\bigg( - cos\bigg[\dfrac{A - B}{2} \bigg] + cos\bigg[\dfrac{A + B}{2} \bigg]\bigg)$

$\rm \:  =  - \: 2sin\bigg[\dfrac{C}{2} \bigg]\bigg(cos\bigg[\dfrac{A - B}{2} \bigg] - cos\bigg[\dfrac{A + B}{2} \bigg]\bigg)$

We know,

$\boxed{ \tt{ \: cos(x - y) - cos(x + y) = 2sinx \: siny \: }}$

So, using this identity, we get

$\rm \:  =  \: - 2sin\bigg[\dfrac{C}{2} \bigg]\bigg(2sin\bigg[\dfrac{A}{2} \bigg]sin\bigg[\dfrac{B}{2} \bigg]\bigg)$

$\rm \:  =  \: - 4 \: sin\bigg[\dfrac{A}{2} \bigg]sin\bigg[\dfrac{B}{2} \bigg]sin\bigg[\dfrac{C}{2} \bigg]$

So, we have

$\boxed{ \sf{sinA+sinB+sinC =- 4 sin\bigg[\dfrac{A}{2} \bigg]sin\bigg[\dfrac{B}{2} \bigg]sin\bigg[\dfrac{C}{2} \bigg]}}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: k \: = \: - \: 4 \: }}$

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Learn More :-

$\boxed{ \tt{ \: sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: cosx - cosy = - 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \tt{ \: 2sinxcosy = sin(x + y) + sin(x - y) \: }}$

$\boxed{ \tt{ \: 2cosx \: cosy = cos(x + y) + cos(x - y) \: }}$