Linearising sin²α cos²α

Problem

Linearise (that is, lower the degree using $\sin^2 = \dfrac{1-\cos(2\,\cdot\,)}{2}$) the expression $\sin^2\alpha\cos^2\alpha$.

Solution

It is convenient first to recognise the product $\sin\alpha\cos\alpha$ as half of $\sin(2\alpha)$: $\sin^2\alpha\cos^2\alpha = (\sin\alpha\cos\alpha)^2 = \left(\frac{1}{2}\sin(2\alpha)\right)^2 = \frac{1}{4}\sin^2(2\alpha).$ Now we apply the linearisation formula to $\sin^2(2\alpha) = \dfrac{1-\cos(4\alpha)}{2}$: $\frac{1}{4}\cdot\frac{1-\cos(4\alpha)}{2} = \boxed{\dfrac{1-\cos(4\alpha)}{8}}.$

Links

Topics: Trigonometric formulae
Concepts: Duplication formulae · Linearisation formulae
Functions: Cosine · Sine
Methods: Duplication formulae
Skills: Proving · Using formulae
Exercise types: Computing expressions