Final answer:
To express the function 3cos(2t)+4sin(2t) as Acos(2t+Q), trigonometric identities and Pythagorean theorem are used to find A and Q. The amplitude A is found to be 5, and the phase shift Q is determined by the arctan function. Superimposition of the original and resultant graphs shows their equivalence.
Step-by-step explanation:
To express the function 3cos(2t)+4sin(2t) in the form Acos(2t+Q), we can use trigonometric identities to combine these two terms into a single trigonometric function that represents a sinusoidal wave with amplitude A and phase shift Q.
We start by recognizing that we're trying to express a sum of a cosine and sine function of the same angular frequency as a single cosine function with an amplitude and phase shift. This is a standard problem in trigonometry and can be solved by using the following identities:
- cos(A + B) = cos(A)cos(B) - sin(A)sin(B)
- sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
Using these identities, we want to find values for A (the amplitude) and Q (the phase shift) that satisfy the equation:
Acos(2t + Q) = Acos(2t)cos(Q) - Asin(2t)sin(Q)
Comparing coefficients, we have the system:
To find A and Q, we can use the Pythagorean identity (since Acos(Q) and -Asin(Q) are the legs of a right triangle with hypotenuse A):
A = sqrt(3^2 + 4^2) = 5
tan(Q) = -4/3
Thus, Q can be found using the arctan function, keeping in mind which quadrant the angle lies in based on the signs of cosine and sine.
Once the values for A and Q are determined, we can plot both the original functions 3cos(2t) and 4sin(2t) and the resulting function 5cos(2t + Q). The graphs can be superimposed to demonstrate that they are equivalent.