The problem is to find A=? B=? and C=?
(7*cos x +3*sin x)^2 = A*cos 2x+B*sin 2x + C
Can anyone help me with this problem?
you didn't say about x. I suppose the problem is in exact form: find the A, B, C so the equation will be true for all x (element of real).
Well, we have three parameter --> we have to get three equation.
I suggest you substitute 3 special value of x in the equation such as x = 0, x=pi/2 and x = pi. Then solve the system of equation for A, B, C.
And in the end you must contol that these values (A, B, C) satisfy indeed the original equation for all x.
(7*cosx+3*sinx)^2 = 49 cos^2 (x) + 9 sin^2 (x) + 2*7*3*Sinx * Cosx
As we know that
Sin2x = 2* sinx * cosx
Cos2x = Cos^2 (x) - Sin^2 (x)
also we know that cos^2 (x) + sin^2 (x) = 1
so solving the above 2 equations we will get
Sin^2 (x) = (1 - cos2x)/2
Cos^2 (x) = (1+cos2x)/2
so putting these values in the above expansion
49 (1 + cos2x)/2 + 9 (1-cos2x)/2 + 21 sin2x
= 29 + 20 cos2x + 21 sin2x
So now u can compare and get the answer
A = 20, B = 21, C = 29