Worked example

Rewrite as a sum: (a) cos 2θ cos 4θ (b) sin θ sin 3θ

Steps

1. cos 2θ cos 4θ = ½[cos(2θ − 4θ) + cos(2θ + 4θ)]
   Cosine-cosine product. · Product to sum: cos A cos B
2. = ½[cos(−2θ) + cos 6θ] = ½[cos 2θ + cos 6θ]
   Cosine is even. · Cosine is even
3. sin θ sin 3θ = ½[cos(θ − 3θ) − cos(θ + 3θ)]
   Sine-sine product. · Product to sum: sin A sin B
4. = ½[cos 2θ − cos 4θ]
   Even cosine again.

Identities used

• Product to sum: cos A cos B — Matching-function products. Set A = B and this is the power-reducing formula cos²A = ½[1 + cos 2A] — which is why it exists in a calculus course.
• Product to sum: sin A sin B — Products of two sines — and, with A = B, the power-reducing formula sin²A = ½[1 − cos 2A]. Also the orthogonality relation Fourier series are built on.

Check it yourself

Open the verifier with the main identity pre-filled — substituting your own numbers into a step is the fastest way to find out which line you did not follow.

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