Using the cosine rule(c^2 =a^2 +b^2 −2abcosC), prove the triangle inequality: if a,b and c are sides of a triangle ABC, then a+b≥c and explain when equality holds. Further prove that sin α + sin β ≥ sin(α+β) for 0° ≤α,β≤180°


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Question Number 112059 by Aina Samuel Temidayo last updated on 05/Sep/20

$Using the cosine$ $rule (c^{2} = a^{2} + b^{2} - 2 abcosC), prove the$ $triangle inequality : if a, b and c are$ $sides of a triangle ABC, then a + b ⩾ c$ $and explain when equality holds .$ $Further prove that \sin α + \sin β ⩾$ $\sin (α + β) for 0 ° ⩽ α, β ⩽ 180 °$
	Answered by 1549442205PVT last updated on 06/Sep/20

	$i) From the cosine theorem we have$ $c^{2} = b^{2} + a^{2} - 2 abcosC$ $\Rightarrow a^{2} + b^{2} + 2 ab = c^{2} + 2 abcosC + 2 ab$ $\Rightarrow {(a + b)}^{2} = c^{2} + 2 ab (1 + cosC) (1)$ $Since C is an angle of the triangle, 0 < C < 180 °$ $\Rightarrow cosC > - 1 \Rightarrow 1 + cosC > 0 . Hence,$ $c^{2} + 2 ab (1 + cosC) > c^{2} (2)$ $From (1) (2) we infer$ ${(a + b)}^{2} > c^{2} \Rightarrow a + b > c$ $ii) We have \sin (α + β) = \sin α \cos β + \cos α \sin β$ $Since 0 < α, β ⩽ 180 °, \sin α, \sin β > 0 . Also,$ $\cos α, \cos β ⩽ 1 . Hence$ $\sin α \cos β ⩽ \sin α (3)$ $\sin β \cos α ⩽ \sin β (4)$ $From (3) (4) we get \sin (α + β) =$ $\sin α \cos β + \cos α \sin β ⩽ \sin α + \sin β (q . e . d)$ $iii) Prove \sin α + \sin β ⩾ 2 \sin \frac{α + β}{2}$ $By the convert formula for the sum$ $of two sine we have$ $\sin α + \sin β = 2 \sin \frac{α + β}{2} \cos \frac{α - β}{2} (5)$ $Since 0 < α < 180, 0 < \frac{α + β}{2} < 90 °$ $\Rightarrow \sin \frac{α + β}{2} > 0 . On the other hands,$ $\cos \frac{α - β}{2} ⩽ 1 . Hence, 2 \sin \frac{α + β}{2} \cos \frac{α + β}{2}$ $⩽ 2 \sin \frac{α + β}{2} (6) . From (5) (6) we get$ $\sin α + \sin β ⩾ 2 \sin \frac{α + β}{2} (q . e . d)$
	Commented by Aina Samuel Temidayo last updated on 06/Sep/20

	$From (1) (2)$ $why do we infer {(a + b)}^{2} > c^{2} . I {don}^{'} t$ $understand please .$
	Commented by Aina Samuel Temidayo last updated on 06/Sep/20

	$For (ii), please how did \sin α + \sin β ⩾$ $\sin (α + β) change into \sin α + \sin β$ $⩾ 2 \sin \frac{α + β}{2}$
	Commented by 1549442205PVT last updated on 06/Sep/20

	$I wrote a mistake and corrected$
	Commented by Aina Samuel Temidayo last updated on 06/Sep/20

	$I {don}^{'} t get . {What}^{'} s the mistake ?$
	Commented by Aina Samuel Temidayo last updated on 06/Sep/20

	$Oh . Thanks . But do I need (iii) ?$
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