1. **Problem statement:** Find the expansions of $\sin(A + B + C)$, $\cos(A + B + C)$, and $\tan(A + B + C)$. 2. **Expand $\sin(A + B + C)$:** Use the angle sum formula for sine: $$\sin(X + Y) = \sin X \cos Y + \cos X \sin Y$$ Let $X = A + B$ and $Y = C$, then $$\sin(A + B + C) = \sin((A + B) + C) = \sin(A + B) \cos C + \cos(A + B) \sin C$$ Now expand $\sin(A + B)$ and $\cos(A + B)$: $$\sin(A + B) = \sin A \cos B + \cos A \sin B$$ $$\cos(A + B) = \cos A \cos B - \sin A \sin B$$ Substitute back: $$\sin(A + B + C) = (\sin A \cos B + \cos A \sin B) \cos C + (\cos A \cos B - \sin A \sin B) \sin C$$ Simplify: $$\sin(A + B + C) = \sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C - \sin A \sin B \sin C$$ 3. **Expand $\cos(A + B + C)$:** Similarly, use the cosine sum formula: $$\cos(X + Y) = \cos X \cos Y - \sin X \sin Y$$ Let $X = A + B$ and $Y = C$, then $$\cos(A + B + C) = \cos(A + B) \cos C - \sin(A + B) \sin C$$ Substitute expansions: $$\cos(A + B) = \cos A \cos B - \sin A \sin B$$ $$\sin(A + B) = \sin A \cos B + \cos A \sin B$$ So, $$\cos(A + B + C) = (\cos A \cos B - \sin A \sin B) \cos C - (\sin A \cos B + \cos A \sin B) \sin C$$ Simplify: $$\cos(A + B + C) = \cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C$$ 4. **Expand $\tan(A + B + C)$:** Use the tangent sum formula for three angles: $$\tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}$$ This formula comes from applying the tangent addition formula twice: $$\tan(X + Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y}$$ with $X = A + B$. **Final answers:** $$\sin(A + B + C) = \sin A \cos B \cos C + \cos A \sin B \cos C + \cos A \cos B \sin C - \sin A \sin B \sin C$$ $$\cos(A + B + C) = \cos A \cos B \cos C - \sin A \sin B \cos C - \sin A \cos B \sin C - \cos A \sin B \sin C$$ $$\tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}$$