The R Formula. How to write an expression in terms of sine or cosine only.
If you have an equation written in terms of sine and cosine, then the R formula can be used to rewrite the equation in terms of sine or cosine only. The 4 forms that you are rewriting the equations in are:
asinѲ + bcosѲ = Rsin(Ѳ + α)
asinѲ - bcosѲ = Rsin(Ѳ - α)
acosѲ +bsinѲ = Rcos (Ѳ - α)
acosѲ -bsinѲ = Rcos (Ѳ + α)
where R is a positive constant and α is a value between 0⁰ and 90⁰. Let’s take a look at a couple examples of using the R formula.
Example 1
Express √3sinѲ – cosѲ in the form Rsin(Ѳ-α), R>0 and 0≤α<90⁰, and solve the equation √3sinѲ – cosѲ = 0.5 in the range 0≤Ѳ<360⁰.
First of all use the addition formula,
sin(A-B) = sinAcosB – cosAsinB,
to expand Rsin(Ѳ-α):
Rsin(Ѳ-α) = R[sinѲcosα – cosѲsinα]
= RsinѲcosα – RcosѲsinα
So we now make this equal to √3sinѲ – cosѲ:
√3sinѲ – cosѲ = RsinѲcosα – RcosѲsinα
Now comparing coefficients on both sides of the equation you can write down 2 new equations:
Rcosα = √3 (Equation A)
Rsinα = 1 (Equation B)
Now if you square both of these equations and add both equations together you can find the value of R:
R²cos²α + R²sin²α = (√3)² + 1²
R²(cos²α + sin²α) = 4
Now you can get rid of the cos²α + sin²α as this is equal to 1 (using the identity cos²A + sin²A = 1)
R² = 4
R = 2
A quick way to work out R, is to square the two numbers from equations A and B, add them together, and take the square root (√[(√3)² + 1²] = √4 = 2 (a method similar to Pythagoras)
Now sub R= 2 back into Rcosα or Rsinα to find α:
2sinα = 1 (divide both sides by 2)
sinα = 0.5 (sin inverse both sides)
α = 30⁰
So √3sinѲ – cosѲ = 2sin(Ѳ-30)
Now we have written this in the R form we can now solve the equation:
2sin(Ѳ-30) = 0.5 (divide both sides by 2)
sin(Ѳ-30) = 0.25 (sin inverse both sides)
Ѳ - 30 = 14.5⁰, 165.5⁰ (make sure you find the other solution in the required range by subtracting the 14.5⁰ from 180⁰, and then add 30⁰ to both solutions)
Ѳ = 44.5⁰ or 195.5⁰.
Lets take a look at one more example of using the R formula:
Example 2
Write 5cosѲ + 3sinѲ in the form R(cosѲ-α).
First of all use the difference formula for cosine to expand R(cosѲ-α) ,
cos(A-B) = cosAcosB + sinAsinB so
Rcos(Ѳ-α) = R[cosѲcosα + sinѲsinα]
= RcosѲcosα + RsinѲsinα
So we now make this equal to 5cosѲ + 3sinѲ:
5cosѲ + 3sinѲ = RcosѲcosα + RsinѲsinα
Now comparing coefficients on both sides of the equation you can write down 2 new equations:
Rcosα = 5 (Equation A)
Rsinα = 3 (Equation B)
Now if you square both of these equations and add both equations together you can find the value of R or you can use the quick Pythagoras method as shown above.
√(5² + 3²) = √34
Now sub R= √34 back into Rcosα = 5 or Rsinα = 3 to find α:
√34sinα = 3 (divide both sides by 3)
sinα = 3/√34 (sin inverse both sides)
α = 31.0⁰
Therefore, 5cosѲ + 3sinѲ = √34(cosѲ-31.0) which you can now go on to use to solve equations.