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√2 cosA=cosB+cos 3 B and √2 sinA=sinB-sin 3 B,then find the value of sin(A-B)

√2 cosA=cosB+cos3B and √2 sinA=sinB-sin3B,then find the value of sin(A-B)
 

Grade:11

1 Answers

Virender Singh
11 Points
5 years ago
√2cosA=cosB+cos3B................(1)
√2sinA=sinB+sin3B.................(2)
 
Multiply (1) with sinB and (2) with cosB
and then subtracting i.e. (1)-(2)
 
=> -√2{ sin(A-B) }= sinBcosB...........(3) 
Now, rewriting (1) as √2cosA-cosB= cos3B and then squaring both sides will => 2cos2A+cos2​B-2√2cosAcosB=cos6B
 
Repeat the process with (2) 
=> 2sin2A+sin2B-2√2sinAsinB=sin6B
 
Add these two resultant eqn. and then solve the resultant as much as possible so that it gets a form like this {note that equation (3) is also used to get to this form}
 
6 cos2(A-B)+ 2√2 cos(A-B) -8=0 
This is quadratic equation in cos(A-B)
 
Solving this will give two values 
cos(A-B)=4/3√2 and
cos(A-B)= -6/3√2
 
Using cos(A-B)=4/3√2
By squaring and converting cos to sin 
we get sin(A-B)= +1/3 and -1/3 
 
 
 

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