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Question Number 18063 by Tinkutara last updated on 14/Jul/17

The angles A, B, C of a triangle ABC satisfy 4cosAcosB + sin2A + sin2B + sin2C = 4. Then which of the following statements is/are correct? (1) The triangle ABC is right angled (2) The triangle ABC is isosceles (3) The triangle ABC is neither isosceles nor right angled (4) The triangle ABC is equilateral

$$\mathrm{The}\:\mathrm{angles}\:{A},\:{B},\:{C}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:{ABC} \\ $$$$\mathrm{satisfy}\:\mathrm{4cos}{A}\mathrm{cos}{B}\:+\:\mathrm{sin2}{A}\:+\:\mathrm{sin2}{B}\:+ \\ $$$$\mathrm{sin2}{C}\:=\:\mathrm{4}.\:\mathrm{Then}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{statements}\:\mathrm{is}/\mathrm{are}\:\mathrm{correct}? \\ $$$$\left(\mathrm{1}\right)\:\mathrm{The}\:\mathrm{triangle}\:{ABC}\:\mathrm{is}\:\mathrm{right}\:\mathrm{angled} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{The}\:\mathrm{triangle}\:{ABC}\:\mathrm{is}\:\mathrm{isosceles} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{The}\:\mathrm{triangle}\:{ABC}\:\mathrm{is}\:\mathrm{neither} \\ $$$$\mathrm{isosceles}\:\mathrm{nor}\:\mathrm{right}\:\mathrm{angled} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{The}\:\mathrm{triangle}\:{ABC}\:\mathrm{is}\:\mathrm{equilateral} \\ $$

Commented by prakash jain last updated on 14/Jul/17

sin 2A+sin 2B+sin 2C =2sin (A+B)cos (A−B)+2sin CcosC =2sin Ccos (A−B)−2sin Ccos(A+B) 4cos Acos B=2cos (A+B)+2cos (A−B) given expression. −cos (A+B)(1−sin C) +cos (A−B)(1+sin C)=2 case 1: if (A+B)<(π/2),sin C<1 cos (A−B)(1+sin C)<2 LHS=2−(some +ve number)≠2 case 2: if (A+B)≥π/2 A+B=(π/2)+x⇒C=(π/2)−x sin x(1−cos x)+cos (A−B)(1+cos x)=2 cos (A−B)=((2−sin x+sin xcos x)/(1+cos x))≥1 RHS=1 iff x=0 cos (A−B)=1⇒A=B ⇒A+B=(π/2)⇒C=(π/2) so triangle is right angles and isoceles.

$$\mathrm{sin}\:\mathrm{2}{A}+\mathrm{sin}\:\mathrm{2}{B}+\mathrm{sin}\:\mathrm{2}{C} \\ $$$$=\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)+\mathrm{2sin}\:{C}\mathrm{cos}{C} \\ $$$$=\mathrm{2sin}\:{C}\mathrm{cos}\:\left({A}−{B}\right)−\mathrm{2sin}\:{C}\mathrm{cos}\left({A}+{B}\right) \\ $$$$\mathrm{4cos}\:{A}\mathrm{cos}\:{B}=\mathrm{2cos}\:\left({A}+{B}\right)+\mathrm{2cos}\:\left({A}−{B}\right) \\ $$$${given}\:{expression}. \\ $$$$−\mathrm{cos}\:\left({A}+{B}\right)\left(\mathrm{1}−\mathrm{sin}\:{C}\right) \\ $$$$\:\:\:\:+\mathrm{cos}\:\left({A}−{B}\right)\left(\mathrm{1}+\mathrm{sin}\:{C}\right)=\mathrm{2} \\ $$$${case}\:\mathrm{1}:\:{if}\:\:\left({A}+{B}\right)<\frac{\pi}{\mathrm{2}},\mathrm{sin}\:{C}<\mathrm{1} \\ $$$$\mathrm{cos}\:\left({A}−{B}\right)\left(\mathrm{1}+\mathrm{sin}\:{C}\right)<\mathrm{2} \\ $$$$\mathrm{LHS}=\mathrm{2}−\left({some}\:+{ve}\:{number}\right)\neq\mathrm{2} \\ $$$${case}\:\mathrm{2}:\:{if}\:\left({A}+{B}\right)\geqslant\pi/\mathrm{2} \\ $$$${A}+{B}=\frac{\pi}{\mathrm{2}}+{x}\Rightarrow{C}=\frac{\pi}{\mathrm{2}}−{x} \\ $$$$\mathrm{sin}\:{x}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)+\mathrm{cos}\:\left({A}−{B}\right)\left(\mathrm{1}+\mathrm{cos}\:{x}\right)=\mathrm{2} \\ $$$$\mathrm{cos}\:\left({A}−{B}\right)=\frac{\mathrm{2}−\mathrm{sin}\:{x}+\mathrm{sin}\:{x}\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}\geqslant\mathrm{1} \\ $$$$\mathrm{RHS}=\mathrm{1}\:{iff}\:{x}=\mathrm{0} \\ $$$$\mathrm{cos}\:\left({A}−{B}\right)=\mathrm{1}\Rightarrow{A}={B} \\ $$$$\Rightarrow{A}+{B}=\frac{\pi}{\mathrm{2}}\Rightarrow{C}=\frac{\pi}{\mathrm{2}} \\ $$$${so}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{right}\:\mathrm{angles}\:\mathrm{and} \\ $$$$\mathrm{isoceles}. \\ $$

Commented by prakash jain last updated on 15/Jul/17

((2−sin x+sin xcos x)/(1+cos x)) =((2−sin x(1−cos x))/(1+cos x)) =((2cos^2 (x/2)+2sin^2 (x/2)−2sin xsin^2 (x/2))/(2cos^2 (x/2))) =1+((sin^2 (x/2)(1−sin x))/(cos^2 (x/2)))≥1

$$\frac{\mathrm{2}−\mathrm{sin}\:{x}+\mathrm{sin}\:{x}\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}} \\ $$$$=\frac{\mathrm{2}−\mathrm{sin}\:{x}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{\mathrm{1}+\mathrm{cos}\:{x}} \\ $$$$=\frac{\mathrm{2cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{2sin}\:{x}\mathrm{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{2cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$$=\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\mathrm{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}\geqslant\mathrm{1} \\ $$

Commented by Tinkutara last updated on 15/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by Tinkutara last updated on 16/Jul/17

In any ΔABC, sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C ∴ 4cosAcosB + 4sinAsinBsinC = 4 sin C = ((1 − cos A cos B)/(sin A sin B)) ≤ 1 ...(i) 1 ≤ cos A cos B + sin A sin B = cos (A − B) ∴ A = B and putting this in (i), sin C = 1 ⇒ C = 90°. Hence ABC is a right angled isosceles triangle.

$$\mathrm{In}\:\mathrm{any}\:\Delta{ABC},\:\mathrm{sin}\:\mathrm{2}{A}\:+\:\mathrm{sin}\:\mathrm{2}{B}\:+ \\ $$$$\mathrm{sin}\:\mathrm{2}{C}\:=\:\mathrm{4}\:\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C} \\ $$$$\therefore\:\mathrm{4cos}{A}\mathrm{cos}{B}\:+\:\mathrm{4sin}{A}\mathrm{sin}{B}\mathrm{sin}{C}\:=\:\mathrm{4} \\ $$$$\mathrm{sin}\:{C}\:=\:\frac{\mathrm{1}\:−\:\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}}{\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}}\:\leqslant\:\mathrm{1}\:...\left(\boldsymbol{{i}}\right) \\ $$$$\mathrm{1}\:\leqslant\:\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}\:+\:\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:=\:\mathrm{cos}\:\left({A}\:−\:{B}\right) \\ $$$$\therefore\:{A}\:=\:{B}\:\mathrm{and}\:\mathrm{putting}\:\mathrm{this}\:\mathrm{in}\:\left(\boldsymbol{{i}}\right), \\ $$$$\mathrm{sin}\:{C}\:=\:\mathrm{1}\:\Rightarrow\:{C}\:=\:\mathrm{90}°. \\ $$$$\mathrm{Hence}\:\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angled}\:\mathrm{isosceles} \\ $$$$\mathrm{triangle}. \\ $$

Commented by prakash jain last updated on 16/Jul/17

Simple approach.

$$\mathrm{Simple}\:\mathrm{approach}. \\ $$

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