The $R \cos(\theta - \alpha)$ method is commonly used in trigonometry to rewrite expressions of the form:

$a \cos\theta + b \sin\theta$

into the single cosine form:

$R \cos(\theta - \alpha)$

where $R$ and $\alpha$ are constants determined using the following steps.

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Step-by-Step Method

1. Compare the Expansion
   Using the cosine angle subtraction identity:

   $\cos(\theta - \alpha) = \cos\theta \cos\alpha + \sin\theta \sin\alpha$

   If we multiply both sides by $R$:

   $R \cos(\theta - \alpha) = R (\cos\alpha \cos\theta + \sin\alpha \sin\theta)$

   Now, compare with $a\cos\theta + b\sin\theta$, giving:

   $a = R \cos\alpha, \quad b = R \sin\alpha$

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2. Find $R$ (Amplitude or Magnitude)
   Use Pythagoras' Theorem to determine $R$:

   $R = \sqrt{a^2 + b^2}$

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3. Find $\alpha$ (Phase Angle)
   Divide the equations:

   $\tan\alpha = \frac{b}{a}$

   Solve for $\alpha$ using:

   $\alpha = \tan^{-1} \left(\frac{b}{a}\right)$

   Make sure to check the correct quadrant based on the signs of $a$ and $b$.

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4. Final Expression
   Substitute $R$ and $\alpha$ back:

   $a\cos\theta + b\sin\theta = R\cos(\theta - \alpha)$

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Example

Express $3\cos\theta + 4\sin\theta$ in the form $R\cos(\theta - \alpha)$.

Step 1: Find $R$

$R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Step 2: Find $\alpha$

$\tan\alpha = \frac{4}{3}$ $\alpha = \tan^{-1} \left(\frac{4}{3}\right)$

Approximating:

$\alpha \approx 53.13^\circ$

Step 3: Final Form

$3\cos\theta + 4\sin\theta = 5\cos(\theta - 53.13^\circ)$

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Related Questions

1. How do you convert $R\cos(\theta - \alpha)$ back to $a\cos\theta + b\sin\theta$?
2. How do you apply the $R \cos(\theta - \alpha)$ method to solve trigonometric equations?
3. Can this method be used in physics (e.g., oscillations, AC circuits)?
4. How does the $R\cos(\theta - \alpha)$ form help in integration problems?
5. What are the geometric interpretations of $R$ and $\alpha$?

Tip:
Always double-check the quadrant when calculating $\alpha$ to ensure the correct angle is used!