Trigonometric Identities from Complex Multiplication

Euler's formula packages cosine and sine into one algebraic object,

$$u(\alpha)=\cos\alpha+i\sin\alpha=e^{i\alpha}.$$

Multiplying these objects lets the real and imaginary coefficients reveal several familiar identities. This is a useful symbolic-computation pattern: perform one polynomial calculation and interpret its coefficients.

Symbolic unit-circle elements

We keep $\alpha$ and $\beta$ symbolic. The expression uMinusBeta represents $u(-\beta)=\cos\beta-i\sin\beta$ without asking the simplifier to know parity rules for sine and cosine.

declare symbol α, β : MathValue

def uAlpha : MathValue := cos α + i * sin α
def uBeta : MathValue := cos β + i * sin β
def uMinusBeta : MathValue := cos β - i * sin β

Angle-addition formulas

The coefficients of $1$ and $i$ in $u(\alpha)u(\beta)$ are, respectively, $\cos(\alpha+\beta)$ and $\sin(\alpha+\beta)$.

coefficients (uAlpha * uBeta) i
$\{-\sin(α) \sin(β) + \cos(α) \cos(β), \cos(β) \sin(α) + \cos(α) \sin(β)\}$

Reading the two returned coefficients gives

$$ \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta, $$ $$ \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta. $$

Adding $u(\alpha)u(\beta)$ and $u(\alpha)u(-\beta)$ isolates the product-to-sum combinations.

coefficients (uAlpha * uBeta + uAlpha * uMinusBeta) i
$\{2 \cos(α) \cos(β), 2 \cos(β) \sin(α)\}$

The result encodes $2\cos\alpha\cos\beta$ in its real part and $2\sin\alpha\cos\beta$ in its imaginary part. Dividing by two and replacing the two products by $u(\alpha\pm\beta)$ yields the standard product-to-sum identities.

Triple-angle formulas

Cubing one unit-circle element performs all terms of the binomial expansion at once.

coefficients (uAlpha ^ 3) i
$\{\cos(α)^{3} - 3 \cos(α) \sin(α)^{2}, -\sin(α)^{3} + 3 \cos(α)^{2} \sin(α)\}$

The real and imaginary components are

$$\cos^3\alpha-3\cos\alpha\sin^2\alpha,$$ $$3\cos^2\alpha\sin\alpha-\sin^3\alpha.$$

Using $\sin^2\alpha=1-\cos^2\alpha$ in the first and $\cos^2\alpha=1-\sin^2\alpha$ in the second gives

$$\cos(3\alpha)=4\cos^3\alpha-3\cos\alpha,$$ $$\sin(3\alpha)=3\sin\alpha-4\sin^3\alpha.$$

The essential point is that Egison's coefficient extraction turns complex multiplication into a transparent derivation, rather than merely checking a pre-stated identity.

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