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

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    Cosine Sine Sum
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Homework Help Overview

The problem involves expressing a sum of the form Asin(α) + Bcos(α) as C sin(α+ϕ). The discussion centers around trigonometric identities and relationships between sine and cosine functions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss expressing cosine in terms of sine and explore the relationships between A, B, C, and ϕ. There are attempts to derive equations relating these variables, including squaring and adding equations to find C.

Discussion Status

The discussion is ongoing with various approaches being explored. Some participants have provided algebraic manipulations and relationships, while others have noted the importance of considering the signs of A and B in determining the quadrant for ϕ.

Contextual Notes

There is a focus on ensuring the correct interpretation of angles and signs, particularly regarding the quadrant in which ϕ lies based on the values of A and B.

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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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