Question

If √2 cosA=cosB+cos^3B & √2sinA=sinB-sin^3B, then proved that sin(A-B) =+-1/3​

Answer 1

√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

hence we can prove it

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