Hello!

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? (Wondering)

Thanks

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- Dec 18th 2008, 03:58 PMKluringenTrigonometry
Hello!

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? (Wondering)

Thanks - Dec 18th 2008, 04:12 PMSkalkaz
Dear Kluringen,

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. - Dec 18th 2008, 04:40 PMsastry.somayajula
(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

and

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