R-Alpha Method

In this article we shall be exploring the R-alpha method. We shall explore how to use expressions such as Rcos(x+α) and Rsin(x-α) in connection with acosx ± bsinx. Furthermore we shall explore how to solve equations of the form acosx ± bsinx=c. You must have some knowledge of the addition formulae to appreciate what is covered here.

Prior knowledge

We use the R-alpha method to rewrite sums or subtraction of two functions into a single function. Take for example trying to sketch the graph of

y = 3sinx + 4cosx

It is quite difficult since the function is the sum of two functions 3sinx and 4cosx.

Rcos(x – α)

Example: Rewrite y = 3sinx + 4cosx into the form y = Rcos(x-α)

Answer:

    Explanation:To write the function we have to find R and α. We’re working with the following functions;

    y = 3sin x + 4cos x
    y = Rcos(x – α)

    We have to compare these two functions to find R and α. First let’s expand y=Rcos(x- α) using the addition formulae.

    y = Rcos(x – α)
    y = R(cosxcosα + sinxsinα)
    y = Rcosxcosα + Rsinxsinα

    Now we compare the two expressions to find the R and α as shown in the following;

    3sin x + 4cos x = Rcosxcosα + Rsinxsinα

    Notice we have to have the RHS matched with the LHS. So we rewrite the RHS to match.

    3sinx + 4cosx = Rsinαsinx + Rcosxcosα

    Be careful here look at what has changed carefully especially the position of the terms in the RHS expression. It’s very important that you understand this because if you don’t you will always fail or get the wrong outcome. It’s very common in students to fail this.

    Now we compare each part

    3sinx + 4cosx = Rsinαsinx + Rcosxcosα
    3sinx = Rsinαsinx
    3 = Rsinα
    3sinx + 4cosx = Rsinαsinx + Rcosxcosα
    4cosx = Rcosxcosα
    4 = Rcosα

    So we have found that;

    3 = Rsinα 1
    4 = Rcosα 2

    To find α we have to divide equation 1 by equation 2 as follows;

    Rsinα/Rcosα = 3/4

    Dividing R by R is one so we take that out;

    sinα/cosα = 3/4

    Sine divided by Cosine is equal to tangent so;

    tanα = 3/4
    α = 36.9°

    To find R we must add equation 1 to equation 2 in this format;

    1² + 2²
    R²sin²α + R²cos²α = 3² + 4²

    …simplify;

    R²(sin²α + cos²α) = 9 + 16

    …we know that sin2α + cos2 α is 1 so;

    R² = 25 which means R = 5

    The function of y = 3sinx + 4cosx when rewritten in Rcos(x- α) becomes;

    3sinx + 4cosx = 5cos(x – 36.9°)

    We have managed to find a simpler function which will enable us to sketch the graph easily;
    [graph id=”28″]

    Example: Write sinx + cosx in the form of Rcos(x- α).

    Answer:

      Explanation:We know that;

      Rcos(x – α) = R(cosxcosα + sinxsinα)

      …from this we get;

      Rsinα = 1 1
      Rcosα = 1 2

      We divide equation 1 by equation 2 to get α as shown below;

      Rsinα/Rcosα = 1/1
      tan α = 1
      α = 45°

      To find R we add the equations in the form of;

      R²sin²α + R²cos²α = 1² + 1²
      R²sin²α + R²cos²α = 1² + 1²
      R²(sin²α + cos²α) = 2
      R = √2

      So now we have;

      sin x + cos x = √2 cos(x – 45°)

      We have managed to rewrite sinx + cosx as √2cos(x-45°).

      Rsin(x + α)

      We can also write two functions as Rsin(x+ α).

      Example: Write 3sinx + 2cosx as Rsin(x+ α)

      Answer:

        Explanation: First we expand Rsin(x+ α) as shown below;

        Rsin(x + α) = R(sinxcosα + cosxsinα)
        = Rsinxcosα + Rcosxsinα

        Then we compare the two functions as follows;

        3sinx + 2cosx = Rsinxcosα + Rcosxsinα

        We rewrite the RHS so that they match with LHS as follows. Take care…

        3sinx + 2cosx = Rsinxcosα + Rcosxsinα

        Now we take out the two equations we are after as we did before;

        3 = Rcosα 1
        2 = Rsinα 2

        To find α we must divide equation 2 by equation 1 as shown below;

        Rsinα/Rcosα = 2/3
        tan α = 2/3
        α = 33.7°

        We have now found α we need to find R by adding equation 1 plus equation 2 as we did before;

        R²sin²α + R²cos²α = 2² + 3²
        R² (sin²α + cos²α) = 4 + 9
        R = √13

        Now we replace the found values into Rsin(x+ α) as follows;

        3sinx + 2cosx = √13sin(x – 33.7°)

        We have managed to rewrite 3sinx + 2cosx as √13sin(x+33.7°). With this now it’s more easy to draw the graph as shown below;

        [GRAPH]

        So now let’s summarise the alpha methods below;

        The above are the alpha methods.

        It would be impossible to solve an equation such as sinx + cosx = 1 because there is no identity to support. The following example shows how to solve equations using the alpha methods.

        Solve equations

        Example: Solve the 4sinx + 6cosx =3 in the range of 0° ≤ x ≤ 360° by dividing the answers to the nearest degree.

        Answer:

          Explanation:
          We have to decide which R-alpha method/formula to use to rewrite 4sinx + 6cosx. We have Rcos(x- α) or Rsin(x+ α) Let’s use Rcos(x- α). First we compare and match as we did above.

          4sinx + 6cosx = Rcos(x – α)
          4sinx + 6cosx = Rsinαsinx – Rcosαcosx

          Notice above I haven’t shown the process of matching but you have to carry out this as it is important.

          We have to find α by diving the equations we have found as shown below;

          Rsinα/Rcosα = 4/6
          tan α = 4/6
          α = 33.7°

          To find R we have to add the equations and finally get;

          R² = 4² + 6²
          R = √52

          Now we replace the α value and R value we have found into Rcos(x- α) to get;

          √52 cos(x – 33.7°) = 3

          because;

          4sinx + 6cosx = √52 cos(x – 33.7°) = 3

          Now we need to find the angles;

          cos(x – 33.7) = 3/√52
          x – 33.7° = 65.4°
          x = 99.1°

          And also;

          x – 33.7° = 360° – 65.4°
          = 294.6°
          x = 294.6 + 33.7°
          x = 328.3°

          The solutions we have found are;

          x = 99° and x = 328°

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