Solution:
f(t)=2sint+3cost
Let f(t)=Asin(t+θ) , where A is amplitude and θ is phase.
So, 2sint+3cost=Asin(t+θ)
⇒2sint+3cost=A[sintcosθ+costsinθ]⇒2sint+3cost=Asintcosθ+Acostsinθ
On comparing both sides,
Acosθ=2,Asinθ=3⇒tanθ=23⇒θ=56.31°
Then, Acosθ=2
⇒Acos56.31°=2⇒A=3.605
Thus, amplitude is 3.605 and phase is 56.31° .
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