Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

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°

$$\mathrm{Using}\:\mathrm{the}\:\mathrm{cosine} \\ $$$$\mathrm{rule}\left(\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2abcosC}\right),\:\mathrm{prove}\:\mathrm{the} \\ $$$$\mathrm{triangle}\:\mathrm{inequality}:\:\mathrm{if}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{are} \\ $$$$\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{ABC},\:\mathrm{then}\:\mathrm{a}+\mathrm{b}\geqslant\mathrm{c} \\ $$$$\mathrm{and}\:\mathrm{explain}\:\mathrm{when}\:\mathrm{equality}\:\mathrm{holds}. \\ $$$$\mathrm{Further}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{sin}\:\alpha\:+\:\mathrm{sin}\:\beta\:\geqslant \\ $$$$\mathrm{sin}\left(\alpha+\beta\right)\:\mathrm{for}\:\mathrm{0}°\:\leqslant\alpha,\beta\leqslant\mathrm{180}° \\ $$

Answered by 1549442205PVT last updated on 06/Sep/20

i)From the cosine theorem we have c^2 =b^2 +a^2 −2abcosC ⇒a^2 +b^2 +2ab=c^2 +2abcosC+2ab ⇒(a+b)^2 =c^2 +2ab(1+cosC)(1) Since C is an angle of the triangle ,0<C<180° ⇒cosC>−1⇒1+cosC>0.Hence, c^2 +2ab(1+cosC)>c^2 (2) From (1)(2) we infer (a+b)^2 >c^2 ⇒a+b>c ii)We have sin(α+β)=sinαcosβ+cosαsinβ Since 0<α,β≤180°,sinα,sinβ>0.Also, cosα,cosβ≤1.Hence sinαcosβ≤sinα(3) sinβcosα≤sinβ (4) From (3)(4)we get sin(α+β)= sinαcosβ+cosαsinβ≤sinα+sinβ(q.e.d) iii)Prove sinα+sinβ≥2sin((α+β)/2) By the convert formula for the sum of two sine we have sinα+sinβ=2sin((α+β)/2)cos((α−β)/2) (5) Since 0<α<180 ,0<((α+β)/2)<90° ⇒sin((α+β)/2)>0 .On the other hands, cos((α−β)/2)≤1.Hence,2sin((α+β)/2)cos((α+β)/2) ≤2sin((α+β)/2) (6).From(5)(6)we get sinα+sinβ≥2sin((α+β)/2) (q.e.d)

$$\left.\mathrm{i}\right)\mathrm{From}\:\mathrm{the}\:\mathrm{cosine}\:\mathrm{theorem}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{c}^{\mathrm{2}} =\mathrm{b}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} −\mathrm{2abcosC} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{2ab}=\mathrm{c}^{\mathrm{2}} +\mathrm{2abcosC}+\mathrm{2ab} \\ $$$$\Rightarrow\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} =\mathrm{c}^{\mathrm{2}} +\mathrm{2ab}\left(\mathrm{1}+\mathrm{cosC}\right)\left(\mathrm{1}\right) \\ $$$$\mathrm{Since}\:\mathrm{C}\:\mathrm{is}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:,\mathrm{0}<\mathrm{C}<\mathrm{180}° \\ $$$$\Rightarrow\mathrm{cosC}>−\mathrm{1}\Rightarrow\mathrm{1}+\mathrm{cosC}>\mathrm{0}.\mathrm{Hence}, \\ $$$$\mathrm{c}^{\mathrm{2}} +\mathrm{2ab}\left(\mathrm{1}+\mathrm{cosC}\right)>\boldsymbol{\mathrm{c}}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{infer}\: \\ $$$$\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} >\mathrm{c}^{\mathrm{2}} \Rightarrow\mathrm{a}+\mathrm{b}>\mathrm{c} \\ $$$$\left.\mathrm{ii}\right)\mathrm{We}\:\mathrm{have}\:\mathrm{sin}\left(\alpha+\beta\right)=\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{cos}\alpha\mathrm{sin}\beta \\ $$$$\mathrm{Since}\:\mathrm{0}<\alpha,\beta\leqslant\mathrm{180}°,\mathrm{sin}\alpha,\mathrm{sin}\beta>\mathrm{0}.\mathrm{Also}, \\ $$$$\mathrm{cos}\alpha,\mathrm{cos}\beta\leqslant\mathrm{1}.\mathrm{Hence} \\ $$$$\mathrm{sin}\alpha\mathrm{cos}\beta\leqslant\mathrm{sin}\alpha\left(\mathrm{3}\right) \\ $$$$\mathrm{sin}\beta\mathrm{cos}\alpha\leqslant\mathrm{sin}\beta\:\left(\mathrm{4}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{3}\right)\left(\mathrm{4}\right)\mathrm{we}\:\mathrm{get}\:\mathrm{sin}\left(\alpha+\beta\right)= \\ $$$$\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{cos}\alpha\mathrm{sin}\beta\leqslant\mathrm{sin}\alpha+\mathrm{sin}\beta\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$$$\left.\mathrm{iii}\right)\mathrm{Prove}\:\mathrm{sin}\alpha+\mathrm{sin}\beta\geqslant\mathrm{2sin}\frac{\alpha+\beta}{\mathrm{2}} \\ $$$$\mathrm{By}\:\mathrm{the}\:\mathrm{convert}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{of}\:\mathrm{two}\:\mathrm{sine}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{sin}\alpha+\mathrm{sin}\beta=\mathrm{2sin}\frac{\alpha+\beta}{\mathrm{2}}\mathrm{cos}\frac{\alpha−\beta}{\mathrm{2}}\:\left(\mathrm{5}\right) \\ $$$$\mathrm{Since}\:\mathrm{0}<\alpha<\mathrm{180}\:,\mathrm{0}<\frac{\alpha+\beta}{\mathrm{2}}<\mathrm{90}° \\ $$$$\Rightarrow\mathrm{sin}\frac{\alpha+\beta}{\mathrm{2}}>\mathrm{0}\:.\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands}, \\ $$$$\mathrm{cos}\frac{\alpha−\beta}{\mathrm{2}}\leqslant\mathrm{1}.\mathrm{Hence},\mathrm{2sin}\frac{\alpha+\beta}{\mathrm{2}}\mathrm{cos}\frac{\alpha+\beta}{\mathrm{2}} \\ $$$$\leqslant\mathrm{2sin}\frac{\alpha+\beta}{\mathrm{2}}\:\left(\mathrm{6}\right).\mathrm{From}\left(\mathrm{5}\right)\left(\mathrm{6}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{sin}\alpha+\mathrm{sin}\beta\geqslant\mathrm{2sin}\frac{\alpha+\beta}{\mathrm{2}}\:\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$

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.

$$\mathrm{From}\:\left(\mathrm{1}\right)\left(\mathrm{2}\right) \\ $$$$\mathrm{why}\:\mathrm{do}\:\mathrm{we}\:\mathrm{infer}\:\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} >\mathrm{c}^{\mathrm{2}} .\:\mathrm{I}\:\mathrm{don}'\mathrm{t} \\ $$$$\mathrm{understand}\:\mathrm{please}. \\ $$

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 ≥2sin((α+β)/2)

$$\mathrm{For}\:\left(\mathrm{ii}\right),\:\mathrm{please}\:\mathrm{how}\:\mathrm{did}\:\mathrm{sin}\:\alpha\:+\:\mathrm{sin}\:\beta\:\geqslant \\ $$$$\mathrm{sin}\left(\alpha+\beta\right)\:\mathrm{change}\:\mathrm{into}\:\mathrm{sin}\:\alpha\:+\:\mathrm{sin}\:\beta \\ $$$$\geqslant\mathrm{2sin}\frac{\alpha+\beta}{\mathrm{2}} \\ $$

Commented by 1549442205PVT last updated on 06/Sep/20

I wrote a mistake and corrected

$$\mathrm{I}\:\mathrm{wrote}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{and}\:\mathrm{corrected} \\ $$

Commented by Aina Samuel Temidayo last updated on 06/Sep/20

I don′t get. What′s the mistake?

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}.\:\mathrm{What}'\mathrm{s}\:\mathrm{the}\:\mathrm{mistake}? \\ $$

Commented by Aina Samuel Temidayo last updated on 06/Sep/20

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

$$\mathrm{Oh}.\:\mathrm{Thanks}.\:\mathrm{But}\:\mathrm{do}\:\mathrm{I}\:\mathrm{need}\:\left(\mathrm{iii}\right)\:? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com