1. **Stating the problem:** Prove or verify the identity: $$\sin^2 \alpha + \sin^2 \beta + \cos^2 (\alpha + \beta) + 2 \sin \alpha \cdot \sin \beta \cdot \cos(\alpha + \beta) = 1$$ 2. **Recall important trigonometric identities:** - Pythagorean identity: $$\sin^2 x + \cos^2 x = 1$$ - Angle sum formulas: $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$ $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$$ 3. **Rewrite the expression grouping terms:** $$\sin^2 \alpha + \sin^2 \beta + \cos^2 (\alpha + \beta) + 2 \sin \alpha \sin \beta \cos(\alpha + \beta)$$ 4. **Use the Pythagorean identity on $$\cos^2(\alpha + \beta)$$:** $$\cos^2(\alpha + \beta) = 1 - \sin^2(\alpha + \beta)$$ 5. **Substitute into the expression:** $$\sin^2 \alpha + \sin^2 \beta + 1 - \sin^2(\alpha + \beta) + 2 \sin \alpha \sin \beta \cos(\alpha + \beta)$$ 6. **Rearrange terms:** $$1 + \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta) + 2 \sin \alpha \sin \beta \cos(\alpha + \beta)$$ 7. **Express $$\sin^2(\alpha + \beta)$$ using the angle sum formula:** $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$ So, $$\sin^2(\alpha + \beta) = (\sin \alpha \cos \beta + \cos \alpha \sin \beta)^2$$ $$= \sin^2 \alpha \cos^2 \beta + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta + \cos^2 \alpha \sin^2 \beta$$ 8. **Substitute $$\sin^2(\alpha + \beta)$$ back:** $$1 + \sin^2 \alpha + \sin^2 \beta - \left(\sin^2 \alpha \cos^2 \beta + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta + \cos^2 \alpha \sin^2 \beta\right) + 2 \sin \alpha \sin \beta \cos(\alpha + \beta)$$ 9. **Recall $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$$ and substitute:** $$2 \sin \alpha \sin \beta (\cos \alpha \cos \beta - \sin \alpha \sin \beta) = 2 \sin \alpha \sin \beta \cos \alpha \cos \beta - 2 \sin^2 \alpha \sin^2 \beta$$ 10. **Rewrite the entire expression:** $$1 + \sin^2 \alpha + \sin^2 \beta - \sin^2 \alpha \cos^2 \beta - 2 \sin \alpha \cos \beta \cos \alpha \sin \beta - \cos^2 \alpha \sin^2 \beta + 2 \sin \alpha \sin \beta \cos \alpha \cos \beta - 2 \sin^2 \alpha \sin^2 \beta$$ 11. **Group like terms:** - Combine $$- 2 \sin \alpha \cos \beta \cos \alpha \sin \beta$$ and $$+ 2 \sin \alpha \sin \beta \cos \alpha \cos \beta$$ which cancel out. 12. **Simplify remaining terms:** $$1 + \sin^2 \alpha + \sin^2 \beta - \sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta - 2 \sin^2 \alpha \sin^2 \beta$$ 13. **Use $$\cos^2 x = 1 - \sin^2 x$$ to rewrite:** $$- \sin^2 \alpha \cos^2 \beta = - \sin^2 \alpha (1 - \sin^2 \beta) = - \sin^2 \alpha + \sin^2 \alpha \sin^2 \beta$$ $$- \cos^2 \alpha \sin^2 \beta = - (1 - \sin^2 \alpha) \sin^2 \beta = - \sin^2 \beta + \sin^2 \alpha \sin^2 \beta$$ 14. **Substitute these back:** $$1 + \sin^2 \alpha + \sin^2 \beta - \sin^2 \alpha + \sin^2 \alpha \sin^2 \beta - \sin^2 \beta + \sin^2 \alpha \sin^2 \beta - 2 \sin^2 \alpha \sin^2 \beta$$ 15. **Simplify terms:** $$1 + (\sin^2 \alpha - \sin^2 \alpha) + (\sin^2 \beta - \sin^2 \beta) + (\sin^2 \alpha \sin^2 \beta + \sin^2 \alpha \sin^2 \beta - 2 \sin^2 \alpha \sin^2 \beta)$$ $$= 1 + 0 + 0 + 0 = 1$$ **Final answer:** $$\sin^2 \alpha + \sin^2 \beta + \cos^2 (\alpha + \beta) + 2 \sin \alpha \sin \beta \cos(\alpha + \beta) = 1$$ This verifies the identity is true.