Example 27

Goal: Prove that \(\cos^3(x)\sin^2(x) = \frac{1}{16}(2\cos(x) - \cos(3x) - \cos(5x))\)

Phase 1: The “Trig Toolbox”

Before starting, a student needs to know these three fundamental identities. Think of these as your building blocks:

  1. Sine Double Angle Identity: \(\sin(2x) = 2\sin(x)\cos(x)\)
    • Rearranged for our use: \(\sin(x)\cos(x) = \frac{1}{2}\sin(2x)\)
  2. Power-Reduction Identity: \(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\)
    • This is used to turn a squared term into a regular term with a double angle.
  3. Product-to-Sum Identity: \(\cos(A)\cos(B) = \frac{1}{2}[\cos(A - B) + \cos(A + B)]\)
    • This is used to turn multiplication into addition/subtraction.

Phase 2: Step-by-Step Proof

Goal: Prove that \(\cos^3(x)\sin^2(x) = \frac{1}{16}(2\cos(x) - \cos(3x) - \cos(5x))\)

Step 1: Group the terms to simplify the powers.

The Left-Hand Side (LHS) is \(\cos^3(x)\sin^2(x)\). We can rewrite this by splitting one \(\cos(x)\) off: \[\text{LHS} = \cos(x) \cdot [\cos^2(x)\sin^2(x)]\] \[\text{LHS} = \cos(x) \cdot [\cos(x)\sin(x)]^2\]

Step 2: Use the Double Angle Identity.

From our toolbox, we know \(\cos(x)\sin(x) = \frac{1}{2}\sin(2x)\). Let’s plug that in: \[\text{LHS} = \cos(x) \cdot \left[ \frac{1}{2}\sin(2x) \right]^2\] Now, square the term inside the brackets: \[\text{LHS} = \cos(x) \cdot \frac{1}{4}\sin^2(2x)\] \[\text{LHS} = \frac{1}{4} \cos(x) \sin^2(2x)\]

Step 3: Reduce the power of \(\sin^2(2x)\).

Using the power-reduction identity from our toolbox (where \(\theta = 2x\)), we get: \(\sin^2(2x) = \frac{1 - \cos(2 \cdot 2x)}{2} = \frac{1 - \cos(4x)}{2}\)

Substitute this back into our expression: \[\text{LHS} = \frac{1}{4} \cos(x) \cdot \left( \frac{1 - \cos(4x)}{2} \right)\] Multiply the fractions (\(\frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}\)): \[\text{LHS} = \frac{1}{8} \cos(x) (1 - \cos(4x))\]

Step 4: Distribute the \(\cos(x)\).

\[\text{LHS} = \frac{1}{8} [\cos(x) - \cos(x)\cos(4x)]\]

Step 5: Apply the Product-to-Sum Identity.

We need to simplify the multiplication term: \(\cos(x)\cos(4x)\). Using our toolbox formula: \(\cos(A)\cos(B) = \frac{1}{2}[\cos(A - B) + \cos(A + B)]\) * Let \(A = x\) and \(B = 4x\): \[\cos(x)\cos(4x) = \frac{1}{2}[\cos(x - 4x) + \cos(x + 4x)]\] \[\cos(x)\cos(4x) = \frac{1}{2}[\cos(-3x) + \cos(5x)]\] * Note for the student: Cosine is an even function, meaning \(\cos(-3x)\) is the same as \(\cos(3x)\). \[\cos(x)\cos(4x) = \frac{1}{2}[\cos(3x) + \cos(5x)]\]

Step 6: Final Substitution and Cleanup.

Put that result back into the main equation from Step 4: \[\text{LHS} = \frac{1}{8} \left[ \cos(x) - \frac{1}{2}(\cos(3x) + \cos(5x)) \right]\]

To make it look like the Right-Hand Side of the original question, we want a fraction of \(\frac{1}{16}\) in the front. To do this, we factor out another \(\frac{1}{2}\) from inside the bracket: \[\text{LHS} = \frac{1}{8} \cdot \frac{1}{2} [2\cos(x) - (\cos(3x) + \cos(5x))]\] \[\text{LHS} = \frac{1}{16} [2\cos(x) - \cos(3x) - \cos(5x)]\]

Conclusion: The Left-Hand Side now exactly matches the Right-Hand Side. \[\text{LHS} = \text{RHS}\] Proved.

NOTE To understand how this rearrangement

1. Start with the “Given” Identity

The standard formula (which you usually find in textbooks) is: \[\sin(2x) = 2 \cdot \sin(x) \cdot \cos(x)\]

2. Use the “Opposite Operation”

\[\frac{\sin(2x)}{\mathbf{2}} = \frac{\mathbf{2} \cdot \sin(x) \cdot \cos(x)}{\mathbf{2}}\]

3. Simplify

Now the equation looks like this: \[\frac{1}{2} \sin(2x) = \sin(x) \cos(x)\]

4. Flip it around

Finally, we just swap the sides so the part we want is on the left: \[\mathbf{\sin(x) \cos(x) = \frac{1}{2} \sin(2x)}\]

Why do we do this? In your main math problem, whenever you see \(\sin(x)\cos(x)\), you can replace it with \(\frac{1}{2}\sin(2x)\). This is a clever trick to turn two different trig functions (sine and cosine) into one single function (sine), which makes the rest of the algebra much easier to handle!