Question

sin square A + sin square b + sin square C is equal to 2 + 2 cos a cos b cos c





$sin {}^{2}a - sin {}^{2}b - sin {}^{2} c$
​

Answer 1

Answer:

Consider the problem,

sin  

2

A+sin  

2

B+sin  

2

c=2+2cosA.cosB.cosc

We can write sin  

2

A as,

sin  

2

A=  

2

1−cos(2A)

​

 

Therefore,

LHS=  

2

1−cos(2A)

​

+  

2

1−cos(2B)

​

+  

2

1−cos(2C)

​

 

=  

2

3

​

−(cos(2A)+cos(2B)+cos(2C))

=  

2

1

​

(3−(2cos(A+B)cos(A−B)+cos(2C)))

C=180−(A+B)

cos(C)=cos(180−(A+B))

cos(C)=−cos(A+B)

​

 

Therefore,

=  

2

3

​

−(−2cos(C)cos(A−B)+cos(2C))

cos(2C)=2cos  

2

(C)−1

​

 

And,

=  

2

1

​

(3−(−2cosC)cos(A−B)+2cos  

2

(C)−1)

=  

2

1

​

(4−(2cos(C)cos(C)−cos(A−B)))

=  

2

1

​

(4−2cos(C)(−cos(A+B)−cos(A−B)))

=  

2

1

​

(4+2cos(C)(cos(A+B)cos(A−B)))

=  

2

1

​

(4+2cos(C)×2cos(A)cos(B))

=2+2cos(A)cos(B)cos(C)

​

 

Therefore, If A+B+C=180, sin  

2

A+sin  

2

B+sin  

2

c=2+2cosA.cosB.cosc