- #1
PainterGuy
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- TL;DR Summary
- I was working on a trigonometric identity and out of curiosity wanted to see how one could go back to the original expression.
Hi,
K₁cos(θt+φ)=K₁cos(θt)cos(φ)-K₁sin(θt)sin(φ)=K₁K₂cos(θt)-K₁K₃sin(θt)
Let's assume φ=30° , K₁=5
5cos(θt+30°) = 5cos(θt)cos(30°)-5sin(θt)sin(30°) = (5)0.866cos(θt)-(5)0.5sin(θt) = 4.33cos(θt)-2.5sin(θt)
If only the final result, 4.33cos(θt)-2.5sin(θt), is given, how do I find the original φ=30° and K₁=5? Or, how do I convert the final result back into 5cos(θt+30°) without knowing any intermediate steps?
Could you please help me with it? Thank you.
Note to self:
sin(A+90°)=sinAcos(90°)+cosAsin90°=cosA therefore cosA=sin(A+90°)
or, cos(A+90°)=cosAcos90°-sinAsin90°=-sinAsin90°
K₁cos(θt+φ)=K₁cos(θt)cos(φ)-K₁sin(θt)sin(φ)=K₁K₂cos(θt)-K₁K₃sin(θt)
Let's assume φ=30° , K₁=5
5cos(θt+30°) = 5cos(θt)cos(30°)-5sin(θt)sin(30°) = (5)0.866cos(θt)-(5)0.5sin(θt) = 4.33cos(θt)-2.5sin(θt)
If only the final result, 4.33cos(θt)-2.5sin(θt), is given, how do I find the original φ=30° and K₁=5? Or, how do I convert the final result back into 5cos(θt+30°) without knowing any intermediate steps?
Could you please help me with it? Thank you.
Note to self:
sin(A+90°)=sinAcos(90°)+cosAsin90°=cosA therefore cosA=sin(A+90°)
or, cos(A+90°)=cosAcos90°-sinAsin90°=-sinAsin90°
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