# Math

posted by jiya

3 cos (ωt + 2) + 4 sin (ωt + 1) as A cos (ωt + φ) .

A = ? φ= ?

1. Steve

3 cos (ωt + 2) = 3(cosωt cos2 - sinωt sin2)
4 sin(ωt+1) = 4(sinωtcos1 + cosωtsin1)

3cosωtcos2+4cosωtsin1
+ 4sinωtcos1 - 3sinωtcos2

Now, according to wolframalpha.com, that gives

A = √(25-24sin(1))
tanφ =

sec(1)(4sin(1)+3cos(2))
-------------------------------
4 - 6sin(1)

so φ ≈ -3.736

See

http://www.wolframalpha.com/input/?i=3+cos+%28t+%2B+2%29+%2B+4+sin+%28t+%2B+1%29

But I don't see any clear path to getting there.

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