If $\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$, show that

(i) $\cos3\alpha+\cos3\beta+\cos3\gamma=3\cos(\alpha+\beta+\gamma)$ and

(ii) $\sin3\alpha+\sin3\beta+\sin3\gamma=3\sin(\alpha+\beta+\gamma)$.

Question

If $\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$, show that

(i) $\cos3\alpha+\cos3\beta+\cos3\gamma=3\cos(\alpha+\beta+\gamma)$ and

(ii) $\sin3\alpha+\sin3\beta+\sin3\gamma=3\sin(\alpha+\beta+\gamma)$.

Accepted Answer

> [!TLDR]
>
> Writing each cosine/sine pair as a point on the unit circle via $\operatorname{cis}$ turns the two given real conditions into one complex condition $x+y+z=0$; the standard algebraic identity for $x+y+z=0$ then produces both trig identities at once by comparing real and imaginary parts.

**Step 1. Introduce the complex numbers $x=\operatorname{cis}\alpha,\ y=\operatorname{cis}\beta,\ z=\operatorname{cis}\gamma$.**
$$x+y+z=(\cos\alpha+\cos\beta+\cos\gamma)+i(\sin\alpha+\sin\beta+\sin\gamma).$$
By hypothesis both the real part and the imaginary part are $0$, so
$$x+y+z=0.$$

**Step 2. Recall the algebraic identity for three numbers summing to zero.** For any $x,y,z$,
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$
Since $x+y+z=0$, the right side vanishes, so
$$x^3+y^3+z^3=3xyz.$$

**Step 3. Compute $x^3,y^3,z^3$ using de Moivre's theorem.**
$$x^3=\operatorname{cis}3\alpha,\quad y^3=\operatorname{cis}3\beta,\quad z^3=\operatorname{cis}3\gamma.$$

**Step 4. Compute $xyz$ using the product rule for arguments.**
$$xyz=\operatorname{cis}\alpha\cdot\operatorname{cis}\beta\cdot\operatorname{cis}\gamma=\operatorname{cis}(\alpha+\beta+\gamma).$$

**Step 5. Substitute into $x^3+y^3+z^3=3xyz$.** …

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Q.If $\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$ , show that

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