Sum of Sine and Cosine: Expressing Any Sum as C sin(α+ϕ)

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SUMMARY

The discussion focuses on expressing the sum of sine and cosine, specifically Asin(α) + Bcos(α), in the form C sin(α + ϕ). The key equations derived include A = C*cos(ϕ) and B = C*sin(ϕ), leading to C = sqrt(A² + B²). The angle ϕ is determined using the formula arctan(B/A), with attention to the signs of A and B to correctly identify the quadrant of ϕ. This transformation is essential for simplifying trigonometric expressions in various mathematical applications.

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Homework Statement


Show that any sum:
Asin(α) + Bcos(α)

can be written as : C sin(α+ϕ)
2. Homework Equations

The Attempt at a Solution


i can express cos(a) as as sin(90-a), and then try to use the formula that adds sines, but it gives the form of cos*sin.
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After expressing C sin(α+ϕ), i managed to get the following:
A = C*cos(phi), B = C*sin(phi),

C = (A+B)/ (Cos(phi) + Sin(phi))
 
Last edited:
Dank2 said:
After expressing C sin(α+ϕ), i managed to get the following:
A = C*cos(phi) (, B = C*sin(phi),

C = (A+B)/ (Cos(phi) + Sin(phi))
One backwards parenthesis .

Once you get A = C⋅cos(ϕ) and B = C⋅sin(ϕ),

Square each equation, then add them.
 
Last edited:
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SammyS said:
One backwards parenthesis .

Once you get A = C⋅cos(ϕ) and B = C⋅sin(ϕ),

Square each equation, then add them.
A^2+B^2 = C2cos2phi + C2sin2phi = C2
C = sqrt(A2+B2)
 
then i can divide B/A = tan (Phi)
and then arctan B/A = Phi

Then we got Csin(α + arctan(B/A)).

Thanks
 
Dank2 said:
then i can divide B/A = tan (Phi)
and then arctan B/A = Phi

Then we got Csin(α + arctan(B/A)).

Thanks
You may need to be careful about the signs of A and B. (Well in this case, just the sign of A.)

If A is negative, then so is cos(ϕ), and that puts ϕ in the 2nd or 3rd quadrant.
 
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