I dont know how I can solve this

Question

The problem is : Reduce the equation to a single cosine function with positive amplitude:

9cos(4πx)−4sin(4πx) = 

−9cos(4πx)−4sin(4πx)= 

4sin(4πx)−9cos(4πx)= 

9cos(4πx)+4sin(4πx)=

Be sure to use at least five decimal places for your answers.

I had put in "9cos(4πx)−4sin(4πx) =(arctan(9/4))/(4pi) ", but the answer was wrong.

Please

Can someone explain to me how it is resolved?

Answer 1

The wording of the problem is confusing, but I will assume you want values for A and φ such that

9cos(4πx) - 4sin(4πx) = Acos(4πx + φ),

with A > 0. Multiply and divide by the square root of the sum of the squares of the coefficients, and distribute the division. Let

4πx = y

√[9² + (-4)²] = √97

9cos(y) - 4sin(y) = √97 [(9 / √97)cos(y) - (4 / √97)sin(y)]

Now use the identity

cos(a + b) = cos(a) cos(b) - sin(a) sin(b)

cos(a) = 9 / √97

sin(a) = 4 / √97

b = y = 4πx

φ = arctan[sin(a) / cos(a)] = arctan(4/9)

a = φ

A = √97

9cos(y) - 4sin(y) = √97 cos(φ + y) = √97 cos(y + φ)

9cos(4πx) - 4sin(4πx) = √97 cos[4πx + arctan(4/9)]

Try working the others in a similar way. You might need to look up different trigonometric identities.