Question

Sina+sinb+sinc=0 cosa+cosb+cosc=0 by complex numbers

Answer 1

Answer:

Step-by-step explanation:

SinA + SinB = -Sinc ---------(1)

CosA + CosB = -CosC ---------(2)

 

Sq and add (1),(2)

2+2Cos(A-B) = 1 ------------(3)

 

Sq and Sub (2),(1)

Cos2A + Cos2B + 2Cos(A+B) = Cos2C

2Cos(A+B)Cos(A-B) + 2Cos(A+B) = Cos2C

or Cos(A+B){2Cos(A-B) + 2} = Cos2C

or Cos(A+B) = Cos2C ----------- [Using (3)]

 

Now SinA+SinB+SinC = 0

Squaring, Sin2A+Sin2B+Sin2C = -2∑SinASinB ----------(4)

 

Now Cos(A+B) = Cos2C

Similarly Cos(B+C) = Cos2A

and Cos(C+A) = Cos2B

So adding these, ∑CosACosB - ∑SinASinB = ∑Cos2A

 

Also Cos(A-B) = -1/2 [from (3)]

similarly Cos(B-C) = -1/2

and Cos(C-A) = -1/2

Adding ∑CosACosB + ∑SinASinB = -3/2

 

So -2∑SinASinB = ∑Cos2A + 3/2

 

Hence from (4) Sin2A+Sin2B+Sin2C = -2∑SinASinB = ∑Cos2A + 3/2

 

Now Cos2A+Cos2B+Cos2C = 2Cos(A+B)Cos(A-B) + Cos2C = -Cos(A+B) + Cos2C = 0 ------ [because Cos(A+B) = Cos2C; and Cos(A-B) = -1/2 which was proved previously]

Hence ∑Cos2A = 0.

 

Hence Sin2A+Sin2B+Sin2C = 3/2.

 

Note that you can have so many results from the above conditions, like

Sin2A+Sin2B+Sin2C = 3/2

∑Cos2A = 0

∑SinASinB = -3/4

∑CosACosB = -3/4

 

Hope this helps.

Answer 2

SinA + SinB = -Sinc ---------(1)

CosA + CosB = -CosC ---------(2)

 

Sq and add (1),(2)

2+2Cos(A-B) = 1 ------------(3)

 

Sq and Sub (2),(1)

Cos2A + Cos2B + 2Cos(A+B) = Cos2C

2Cos(A+B)Cos(A-B) + 2Cos(A+B) = Cos2C

or Cos(A+B){2Cos(A-B) + 2} = Cos2C

or Cos(A+B) = Cos2C ----------- [Using (3)]

 

Now SinA+SinB+SinC = 0

Squaring, Sin2A+Sin2B+Sin2C = -2∑SinASinB ----------(4)

 

Now Cos(A+B) = Cos2C

Similarly Cos(B+C) = Cos2A

and Cos(C+A) = Cos2B

So adding these, ∑CosACosB - ∑SinASinB = ∑Cos2A

 

Also Cos(A-B) = -1/2 [from (3)]

similarly Cos(B-C) = -1/2

and Cos(C-A) = -1/2

Adding ∑CosACosB + ∑SinASinB = -3/2

 

So -2∑SinASinB = ∑Cos2A + 3/2

 

Hence from (4) Sin2A+Sin2B+Sin2C = -2∑SinASinB = ∑Cos2A + 3/2

 

Now Cos2A+Cos2B+Cos2C = 2Cos(A+B)Cos(A-B) + Cos2C = -Cos(A+B) + Cos2C = 0 ------ [because Cos(A+B) = Cos2C; and Cos(A-B) = -1/2 which was proved previously]

Hence ∑Cos2A = 0.

 

Hence Sin2A+Sin2B+Sin2C = 3/2.

 

Note that you can have so many results from the above conditions, like

Sin2A+Sin2B+Sin2C = 3/2

∑Cos2A = 0

∑SinASinB = -3/4

∑CosACosB = -3/4

 
$hope \: that \: this \: answer \: will \: help \: u$