# Addition of Sine & Cosine

1. Feb 19, 2013

### fonz

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

R = √a2+b2
α = arctan(b/a)

3. 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 cant see how this is necessary).

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

2. Feb 19, 2013

### vela

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

3. Feb 19, 2013

### fonz

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)

Last edited: Feb 19, 2013
4. Feb 19, 2013

### vela

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

5. Feb 19, 2013

### vela

Staff Emeritus
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.