Question

$if, \: \: \:\:\: \cos \: x \: + \cos \: y = a \\ \: \: \: and \:\: \:\: \sin \: x + \sin \: y \: = b \\ then \: prove \: that, \: \: \\ \cos(x + y) = \frac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} }$

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Answer 1

Given :  cos x + cosy  = a

sinx + siny  = b

To Find : Prove that Cos(x + y) = (a² -  b²   )/  (a² + b²)

Solution:

cos x + cosy  = a

=> cos²x + cos² y  + 2cosxcosy   = a²   Eq1

sinx + siny  = b

=> sin²x + sin² y  + 2sinxsiny   = b²   Eq2

Adding both

1 + 1 + 2(cosxcosy + sinxsiny) = a² + b²

=> 2 ( 1 + cos(x - y) ) =  a² + b²    Eq3

Eq1 - Eq2

=> cos²x  - sin²x + cos² y -  sin² y+ 2cosxcosy   -  2sinxsiny   =a² -  b²    

=> cos2x  + cos2y  + 2(cos(x + y)) = a² -  b²  

=> 2Cos(x + y)cos(x - y)  +  2(cos(x + y)) = a² -  b²  

=>  2Cos(x + y) (Cos(x - y)  + 1)  = a² -  b²  

=> Cos(x + y)  * 2(1 + cos(x - y) )  =  a² -  b²  

from Eq3

=> Cos(x + y) ( a² + b²)  =  a² -  b²  

=> Cos(x + y) = (a² -  b²   )/  (a² + b²)

QED

Hence proved

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Answer 2

Answer:

Given :  cos x + cosy  = a

sinx + siny  = b

To Find : Prove that Cos(x + y) = (a² -  b²   )/  (a² + b²)

Solution:

cos x + cosy  = a

=> cos²x + cos² y  + 2cosxcosy   = a²   Eq1

sinx + siny  = b

=> sin²x + sin² y  + 2sinxsiny   = b²   Eq2

Adding both

1 + 1 + 2(cosxcosy + sinxsiny) = a² + b²

=> 2 ( 1 + cos(x - y) ) =  a² + b²    Eq3

Eq1 - Eq2

=> cos²x  - sin²x + cos² y -  sin² y+ 2cosxcosy   -  2sinxsiny   =a² -  b²    

=> cos2x  + cos2y  + 2(cos(x + y)) = a² -  b²  

=> 2Cos(x + y)cos(x - y)  +  2(cos(x + y)) = a² -  b²  

=>  2Cos(x + y) (Cos(x - y)  + 1)  = a² -  b²  

=> Cos(x + y)  * 2(1 + cos(x - y) )  =  a² -  b²  

from Eq3

=> Cos(x + y) ( a² + b²)  =  a² -  b²  

=> Cos(x + y) = (a² -  b²   )/  (a² + b²)

QED

Hence proved

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if A cos - B sin = C , prove that A sin + B cos = +- whole root (A^2 + B ...

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