Write −6sin(8t)+5cos(8t) in the form Asin(Bt+ϕ) using sum or difference formulas. −6sin(8t)+5cos(8t) =

Answer 1

Using the sum formula for sine, which states that sin(a+b) = sin(a)cos(b) + cos(a)sin(b):

Asin(Bt + φ) = A(sin(Bt)cos(φ) + cos(Bt)sin(φ)) = Asin(Bt)cos(φ) + Acos(Bt)sin(φ) = [Acos(φ)]sin(Bt) + [Asin(φ)]cos(Bt)

Since we are to write -6sin(8t) + 5cos(8t) in this form:

-6sin(8t) + 5cos(8t) = Asin(Bt + φ) = [Acos(φ)]sin(Bt) + [Asin(φ)]cos(Bt)

-6sin(8t) + 5cos(8t) = [Acos(φ)]sin(Bt) + [Asin(φ)]cos(Bt)

We can now see that to write the left side of the equation as the right side, the following must be true:

Acos(φ) = -6, Asin(φ) = 5, B = 8

We now need to solve for A and φ!

We can solve for φ by solving for tan(φ) and cancelling out A as follows:

Asin(φ)/Acos(φ) = 5/-6

sin(φ)/cos(φ) = -5/6

tan(φ) = -5/6

φ = tan^-1(-5/6) = -39.81°

Edit: As Doug C. points out, since A is positive and based on what we know about Asin(φ) and Acos(φ), φ should be in Quadrant II.

To get there from Quadrant IV, we add by 180° to get 140.19°.

Therefore, φ = 140.19°.

Now we can solve for A by using the Pythagorean formula (sin²Θ + cos²Θ= 1).

[Asin(φ)]² + [Acos(φ)]² = 5² + (-6)²

A²sin²(φ) + A²cos²(φ) = 25 + 36

A²[sin²(φ) + cos²(φ)] = 61

A² = 61

A = √(61)

By plugging in A, B, and φ, we can arrive at our final answer:

-6sin(8t) + 5cos(8t) = √(61)sin(8t + 140.19°)