Question

In any traingle,ABC,show that: $\tt{{ \sin}^{2} \frac{A}{2} + {\sin} ^{2} \frac{B}{2} + { \sin}^{2} \frac{C}{2}= 1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}$​

Answer 1

To Prove :-

$\tt{{ \sin}^{2} \frac{A}{2} + {\sin} ^{2} \frac{B}{2} + { \sin}^{2} \frac{C}{2}= 1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}$

Identities Used :-

$\boxed{ \bf{ \: {2sin}^{2}x = 1 - cos2x}}$

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

$\boxed{ \bf{ \: 1 - {2sin}^{2}x = cos2x}}$

$\boxed{ \bf{ \: cos(x - y) - cos(x + y) = 2sinxsiny}}$

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

Consider LHS

$\rm :\longmapsto\:\tt{{ \sin}^{2} \dfrac{A}{2} + {\sin} ^{2} \dfrac{B}{2} + { \sin}^{2} \dfrac{C}{2}}$

On multiply and divide by 2, we get

$\rm \:  = \dfrac{1}{2}( \:\tt{{2 \sin}^{2} \dfrac{A}{2} +2 {\sin} ^{2} \dfrac{B}{2} + 2{ \sin}^{2} \dfrac{C}{2}})$

Using the identity,

$\boxed{ \bf{ \: {2sin}^{2}x = 1 - cos2x}}$

we get

$\rm \:  =  \:  \: \dfrac{1}{2}(1 - cosA + 1 - cosB + 1 - cosC)$

$\rm \:  =  \:  \: \dfrac{1}{2}(3 - cosA - cosB - cosC)$

can be rewritten as

$\rm \:  =  \:  \: \dfrac{1}{2}[3 - (cosA + cosB + cosC)] - - (1)$

Let we first solve

$\rm :\longmapsto\:cosA + cosB + cosC$

Using the identity,

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

we get

$\rm \:  =  \:  \: 2cos\bigg(\dfrac{A + B}{2} \bigg)cos\bigg(\dfrac{A - B}{2} \bigg) + cosC$

As

$\boxed{ \bf{ \: A + B + C = \pi}}$

$\rm \:  =  \:  \: 2cos\bigg(\dfrac{\pi - C}{2} \bigg)cos\bigg(\dfrac{A - B}{2} \bigg) + cosC$

$\rm \:  =  \:  \: 2cos\bigg(\dfrac{\pi}{2} - \dfrac{C}{2} \bigg)cos\bigg(\dfrac{A - B}{2} \bigg) + cosC$

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

We know,

$\boxed{ \bf{ \: 1 - {2sin}^{2}x = cos2x}}$

we get

$\rm \:  =  \:  \: 2sin\bigg(\dfrac{C}{2} \bigg)cos\bigg(\dfrac{A - B}{2} \bigg) + 1 - 2 {sin}^{2} \dfrac{C}{2}$

$\rm \:  =  \:  \:1 + 2sin\bigg(\dfrac{C}{2} \bigg)cos\bigg(\dfrac{A - B}{2} \bigg) - 2 {sin}^{2} \dfrac{C}{2}$

$\rm \:  =  \:  \:1 + 2sin\dfrac{C}{2}\bigg[ cos\bigg(\dfrac{A - B}{2} \bigg) - {sin}^{} \dfrac{C}{2}\bigg]$

$\rm \:  = \:1 + 2sin\dfrac{C}{2}\bigg[ cos\bigg(\dfrac{A - B}{2} \bigg) - {sin}^{} \dfrac{\pi - (A + B)}{2}\bigg]$

$\rm \:  = \:1 + 2sin\dfrac{C}{2}\bigg[ cos\bigg(\dfrac{A - B}{2} \bigg) - {sin}^{}\bigg(\dfrac{\pi}{2} - \dfrac{(A + B)}{2}\bigg)\bigg]$

$\rm \:  = \:1 + 2sin\dfrac{C}{2}\bigg[ cos\bigg(\dfrac{A - B}{2} \bigg) - {cos}^{}\bigg( \dfrac{(A + B)}{2}\bigg)\bigg]$

We know,

$\boxed{ \bf{ \: cos(x - y) - cos(x + y) = 2sinxsiny}}$

So, we get

$\rm \:  =  \:  \: 1 + 2sin\bigg(\dfrac{C}{2} \bigg) \times 2sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)$

$\rm \:  =  \:  \: 1 + 4sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)sin\bigg(\dfrac{C}{2} \bigg) - - - (2)$

On substituting the value of equation (2) in equation (1),

$\rm \:  =  \:  \: \dfrac{1}{2}\bigg[3 -1 - 4sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)sin\bigg(\dfrac{C}{2} \bigg)\bigg ]$

$\rm \:  =  \:  \: \dfrac{1}{2}\bigg[2 - 4sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)sin\bigg(\dfrac{C}{2} \bigg)\bigg ]$

$\rm \:  =  \:  \:1 - 2sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)sin\bigg(\dfrac{C}{2} \bigg)$

Hence,

$\rm :\longmapsto\:\tt{{ \sin}^{2} \dfrac{A}{2} + {\sin} ^{2} \dfrac{B}{2} + { \sin}^{2} \dfrac{C}{2}}$

$\rm \:  =  \:  \:1 - 2sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)sin\bigg(\dfrac{C}{2} \bigg)$

Answer 2

Answer:

To Prove :-

\tt{{ \sin}^{2} \frac{A}{2} + {\sin} ^{2} \frac{B}{2} + { \sin}^{2} \frac{C}{2}= 1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}sin

2

2

A

+sin

2

2

B

+sin

2

2

C

=1−2sin

2

A

sin

2

B

sin

2

C

Identities Used :-

\boxed{ \bf{ \: {2sin}^{2}x = 1 - cos2x}}

2sin

2

x=1−cos2x

\boxed{ \bf{ \: cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)}}

cosx+cosy=2cos(

2

x+y

)cos(

2

x−y

)

\boxed{ \bf{ \: 1 - {2sin}^{2}x = cos2x}}

1−2sin

2

x=cos2x

\boxed{ \bf{ \: cos(x - y) - cos(x + y) = 2sinxsiny}}

cos(x−y)−cos(x+y)=2sinxsiny

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

Solution−

Consider LHS

\rm :\longmapsto\:\tt{{ \sin}^{2} \dfrac{A}{2} + {\sin} ^{2} \dfrac{B}{2} + { \sin}^{2} \dfrac{C}{2}}:⟼sin

2

2

A

+sin

2

2

B

+sin

2

2

C

On multiply and divide by 2, we get

\rm \: = \dfrac{1}{2}( \:\tt{{2 \sin}^{2} \dfrac{A}{2} +2 {\sin} ^{2} \dfrac{B}{2} + 2{ \sin}^{2} \dfrac{C}{2}}) =

2

1

(2sin

2

2

A

+2sin

2

2

B

+2sin

2

2

C

)

Using the identity,

\boxed{ \bf{ \: {2sin}^{2}x = 1 - cos2x}}

2sin

2

x=1−cos2x

we get

\rm \: = \: \: \dfrac{1}{2}(1 - cosA + 1 - cosB + 1 - cosC) =

2

1

(1−cosA+1−cosB+1−cosC)

\rm \: = \: \: \dfrac{1}{2}(3 - cosA - cosB - cosC) =

2

1

(3−cosA−cosB−cosC)

can be rewritten as

\rm \: = \: \: \dfrac{1}{2}[3 - (cosA + cosB + cosC)] - - (1) =

2

1

[3−(cosA+cosB+cosC)]−−(1)

Let we first solve

\rm :\longmapsto\:cosA + cosB + cosC:⟼cosA+cosB+cosC

Using the identity,

\boxed{ \bf{ \: cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)}}

cosx+cosy=2cos(

2

x+y

)cos(

2

x−y

)

we get

\rm \: = \: \: 2cos\bigg(\dfrac{A + B}{2} \bigg)cos\bigg(\dfrac{A - B}{2} \bigg) + cosC = 2cos(

2

A+B

)cos(

2

A−B

)+cosC

As

\boxed{ \bf{ \: A + B + C = \pi}}

A+B+C=π

\rm \: = \: \: 2cos\bigg(\dfrac{\pi - C}{2} \bigg)cos\bigg(\dfrac{A - B}{2} \bigg) + cosC = 2cos(

2

π−C

)cos(

2

A−B

)+cosC

\rm \: = \: \: 2cos\bigg(\dfrac{\pi}{2} - \dfrac{C}{2} \bigg)cos\bigg(\dfrac{A - B}{2} \bigg) + cosC = 2cos(

2

π

−

2

C

)cos(

2

A−B

)+cosC

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

2

C

)cos(

2

A−B

)+cosC

We know,

\boxed{ \bf{ \: 1 - {2sin}^{2}x = cos2x}}

1−2sin

2

x=cos2x

we get

\rm \: = \: \: 2sin\bigg(\dfrac{C}{2} \bigg)cos\bigg(\dfrac{A - B}{2} \bigg) + 1 - 2 {sin}^{2} \dfrac{C}{2} = 2sin(

2

C

)cos(

2

A−B

)+1−2sin

2

2

C

\rm \: = \: \:1 + 2sin\bigg(\dfrac{C}{2} \bigg)cos\bigg(\dfrac{A - B}{2} \bigg) - 2 {sin}^{2} \dfrac{C}{2} = 1+2sin(

2

C

)cos(

2

A−B

)−2sin

2

2

C

\rm \: = \: \:1 + 2sin\dfrac{C}{2}\bigg[ cos\bigg(\dfrac{A - B}{2} \bigg) - {sin}^{} \dfrac{C}{2}\bigg] = 1+2sin

2

C

[cos(

2

A−B

)−sin

2

C

]

\rm \: = \:1 + 2sin\dfrac{C}{2}\bigg[ cos\bigg(\dfrac{A - B}{2} \bigg) - {sin}^{} \dfrac{\pi - (A + B)}{2}\bigg] = 1+2sin

2

C

[cos(

2

A−B

)−sin

2

π−(A+B)

]

\rm \: = \:1 + 2sin\dfrac{C}{2}\bigg[ cos\bigg(\dfrac{A - B}{2} \bigg) - {sin}^{}\bigg(\dfrac{\pi}{2} - \dfrac{(A + B)}{2}\bigg)\bigg] = 1+2sin

2

C

[cos(

2

A−B

)−sin

(

2

π

−

2

(A+B)

)]

\rm \: = \:1 + 2sin\dfrac{C}{2}\bigg[ cos\bigg(\dfrac{A - B}{2} \bigg) - {cos}^{}\bigg( \dfrac{(A + B)}{2}\bigg)\bigg] = 1+2sin

2

C

[cos(

2

A−B

)−cos

(

2

(A+B)

)]

We know,

\boxed{ \bf{ \: cos(x - y) - cos(x + y) = 2sinxsiny}}

cos(x−y)−cos(x+y)=2sinxsiny

So, we get

\rm \: = \: \: 1 + 2sin\bigg(\dfrac{C}{2} \bigg) \times 2sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg) = 1+2sin(

2

C

)×2sin(

2

A

)sin(

2

B

)

\rm \: = \: \: 1 + 4sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)sin\bigg(\dfrac{C}{2} \bigg) - - - (2) = 1+4sin(

2

A

)sin(

2

B

)sin(

2

C

)−−−(2)

On substituting the value of equation (2) in equation (1),

\rm \: = \: \: \dfrac{1}{2}\bigg[3 -1 - 4sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)sin\bigg(\dfrac{C}{2} \bigg)\bigg ] =

2

1

[3−1−4sin(

2

A

)sin(

2

B

)sin(

2

C

)]

\rm \: = \: \: \dfrac{1}{2}\bigg[2 - 4sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)sin\bigg(\dfrac{C}{2} \bigg)\bigg ] =

2

1

[2−4sin(

2

A

)sin(

2

B

)sin(

2

C

)]

\rm \: = \: \:1 - 2sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)sin\bigg(\dfrac{C}{2} \bigg) = 1−2sin(

2

A

)sin(

2

B

)sin(

2

C

)

Hence,

\rm :\longmapsto\:\tt{{ \sin}^{2} \dfrac{A}{2} + {\sin} ^{2} \dfrac{B}{2} + { \sin}^{2} \dfrac{C}{2}}:⟼sin

2

2

A

+sin

2

2

B

+sin

2

2

C

\rm \: = \: \:1 - 2sin\bigg(\dfrac{A}{2} \bigg)sin\bigg(\dfrac{B}{2} \bigg)sin\bigg(\dfrac{C}{2} \bigg) = 1−2sin(

2

A

)sin(

2

B

)sin(

2

C

)