√2 cosA=cosB+cos3B and √2 sinA=sinB-sin3B,then find the value of sin(A-B)
Anubhab , 8 Years ago
Grade 11
Trigonometry> √2 cosA=cosB+cos 3 B and √2 sinA=sinB-sin...
1 AnswersVirender Singh
Last Activity: 8 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+cos2B-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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