How Do You Combine 3sin(ωt) and 2cos(ωt) into a Single Trigonometric Function?

[fonz]

Homework Statement

Express 3sin(ωt) + 2cos(ωt) in the form Rsin(ωt + α)

AND

verify the resultant function is the same frequency as 3sin(ωt) and 2cos(ωt)

Homework Equations

R = √a²+b²
α = arctan(b/a)

The Attempt at a Solution

My attempt using the equations above produces the answer R = √13 and α = 0.6 rad or 33.7°

My argument for the solution being the same frequency is that the period T = 2∏/ω in each case therefore f = 1/T = ω/2∏

I have my doubts about this solution because I believe the marks are awarded for a solution involving the double and compound angle formula (I can't see how this is necessary).

Also, is the statement regarding frequency comprehensive enough or is there a better way of presenting the solution?

 

[vela]

Do you know how to go from the first form to the second form other than by plugging numbers into the two formulas?

 

[fonz]

Do you know how to go from the first form to the second form other than by plugging numbers into the two formulas?

R is the hypotenuse of the right angle triangle with base a and height b. Therefore R² = a² + b² (pythagoras).

the angle can be solved by finding the arctan of b/a (basic trig).

This is not the question though?

EDIT: Solution with compound angle formula

√13sin(ωt+0.6) = √13(sin(ωt)cos(0.6) + cos(ωt)sin(0.6))

= √13(0.83sin(ωt) + 0.56cos(ωt))

= 3sin(ωt) + 2cos(ωt)

how does this prove the frequency being the same? (even though I know it is the same)

 

[vela]

That's right, but how does that relate to the fact that $a \sin \omega t + b \cos \omega t = R \sin(\omega t+\alpha)$?

 

[vela]

I'm not sure exactly what your instructor is looking for regarding the frequency. As you stated the problem, it's already assumed the frequencies are equal.