Question Number 14269 by myintkhaing last updated on 30/May/17
$${In}\:\Delta{ABC},\:{if}\:{sinA}+{sinB}+{sinC}.={cosA}+{cosB}+{cosC}+\mathrm{1}, \\ $$$${then}\:{prove}\:{that}\:\Delta{ABC}\:{is}\:{a}\:{right}\:{triangle}. \\ $$
Answered by prakash jain last updated on 30/May/17
![sin A−cos A+sin B−cos B+sin C−cos C=1 cos A=sin ((π/2)−A) sin A−sin ((π/2)−A)=2cos ((π/4))sin (A−(π/4)) =(√2)sin (A−(π/4)) sin (A−(π/4))+sin (B−(π/4))+sin (C−(π/4))=(1/(√2))=sin ((π/4)) sin (A−(π/4))+sin (B−(π/4))=sin (π/4)−sin (C−(π/4)) 2sin (((A+B)/2)−(π/4))cos (((A−B)/2))=2cos (C/2)sin ((π/4)−(C/2)) 2sin (((π−C)/2)−(π/4))cos (((A−B)/2))=2cos (C/2)sin ((π/4)−(C/2)) sin ((π/4)−(C/2))cos (((A−B)/2))=cos (C/2)sin ((π/4)−(C/2)) sin ((π/4)−(C/2))[cos (((A−B)/2))−cos (C/2)]=0 either case 1 sin ((π/4)−(C/2))=0⇒C=(π/2) or cos (((A−B)/2))=cos (C/2) case 2: (C/2)=((A−B)/2)⇒π−A−B=A−B⇒A=(π/2) case 3: (C/2)=2π−((A−B)/2) ((π−A−B)/2)=((4π−A+B)/2) B=−((3π)/2)⇒B=(π/2) In all the cases one of the angles=(π/2)■](Q14337.png)
$$\mathrm{sin}\:\mathrm{A}−\mathrm{cos}\:\mathrm{A}+\mathrm{sin}\:\mathrm{B}−\mathrm{cos}\:\mathrm{B}+\mathrm{sin}\:{C}−\mathrm{cos}\:{C}=\mathrm{1} \\ $$$$\mathrm{cos}\:{A}=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{A}\right) \\ $$$$\mathrm{sin}\:{A}−\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{A}\right)=\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{4}}\right)\mathrm{sin}\:\left({A}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\sqrt{\mathrm{2}}\mathrm{sin}\:\left({A}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{sin}\:\left({A}−\frac{\pi}{\mathrm{4}}\right)+\mathrm{sin}\:\left({B}−\frac{\pi}{\mathrm{4}}\right)+\mathrm{sin}\:\left({C}−\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{sin}\:\left({A}−\frac{\pi}{\mathrm{4}}\right)+\mathrm{sin}\:\left({B}−\frac{\pi}{\mathrm{4}}\right)=\mathrm{sin}\:\frac{\pi}{\mathrm{4}}−\mathrm{sin}\:\left({C}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{2sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)=\mathrm{2cos}\:\frac{{C}}{\mathrm{2}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{{C}}{\mathrm{2}}\right) \\ $$$$\mathrm{2sin}\:\left(\frac{\pi−{C}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)=\mathrm{2cos}\:\frac{{C}}{\mathrm{2}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{{C}}{\mathrm{2}}\right) \\ $$$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{{C}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)=\mathrm{cos}\:\frac{{C}}{\mathrm{2}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{{C}}{\mathrm{2}}\right) \\ $$$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{{C}}{\mathrm{2}}\right)\left[\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)−\mathrm{cos}\:\frac{{C}}{\mathrm{2}}\right]=\mathrm{0} \\ $$$${either}\:\:\mathrm{case}\:\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{{C}}{\mathrm{2}}\right)=\mathrm{0}\Rightarrow{C}=\frac{\pi}{\mathrm{2}} \\ $$$${or} \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)=\mathrm{cos}\:\frac{{C}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{case}\:\mathrm{2}:\:\frac{{C}}{\mathrm{2}}=\frac{{A}−{B}}{\mathrm{2}}\Rightarrow\pi−{A}−{B}={A}−{B}\Rightarrow{A}=\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{case}\:\mathrm{3}: \\ $$$$\:\:\:\:\:\:\:\:\frac{{C}}{\mathrm{2}}=\mathrm{2}\pi−\frac{{A}−{B}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\pi−{A}−{B}}{\mathrm{2}}=\frac{\mathrm{4}\pi−{A}+{B}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{B}=−\frac{\mathrm{3}\pi}{\mathrm{2}}\Rightarrow{B}=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{In}\:\mathrm{all}\:\mathrm{the}\:\mathrm{cases}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angles}=\frac{\pi}{\mathrm{2}}\blacksquare \\ $$