Question-68210

Question Number 68210 by peter frank last updated on 07/Sep/19

Answered by $@ty@m123 last updated on 08/Sep/19

Let ((sin A)/a)=((sin B)/b)=((sin C)/c)=R ⇒sin A=aR, sin B=bR, sin C=cR ...(1) (2) ((sin (A−B)sin C)/(1+cos (A−B)cos C)) = ((sin (A−B)sin (A+B))/(1−cos (A−B)cos (A+B))) =((sin^2 A−sin^2 B)/(1−cos^2 A+sin^2 B)) =((sin^2 A−sin^2 B)/(sin ^2 A+sin^2 B)) =((a^2 −b^2 )/(a^2 +b^2 )) , using (1)

$${Let}\:\frac{\mathrm{sin}\:{A}}{{a}}=\frac{\mathrm{sin}\:{B}}{{b}}=\frac{\mathrm{sin}\:{C}}{{c}}={R} \\ $$$$\Rightarrow\mathrm{sin}\:{A}={aR},\:\mathrm{sin}\:{B}={bR},\:\mathrm{sin}\:{C}={cR}\:…\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{sin}\:\left({A}−{B}\right)\mathrm{sin}\:{C}}{\mathrm{1}+\mathrm{cos}\:\left({A}−{B}\right)\mathrm{cos}\:{C}} \\ $$$$=\:\frac{\mathrm{sin}\:\left({A}−{B}\right)\mathrm{sin}\:\left({A}+{B}\right)}{\mathrm{1}−\mathrm{cos}\:\left({A}−{B}\right)\mathrm{cos}\:\left({A}+{B}\right)} \\ $$$$=\frac{\mathrm{sin}\:^{\mathrm{2}} {A}−\mathrm{sin}\:^{\mathrm{2}} {B}}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {A}+\mathrm{sin}\:^{\mathrm{2}} {B}} \\ $$$$=\frac{\mathrm{sin}\:^{\mathrm{2}} {A}−\mathrm{sin}\:^{\mathrm{2}} {B}}{\mathrm{sin}\:\:^{\mathrm{2}} {A}+\mathrm{sin}\:^{\mathrm{2}} {B}} \\ $$$$=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:,\:{using}\:\left(\mathrm{1}\right) \\ $$

Commented by peter frank last updated on 08/Sep/19

$${thanks}\:{for}\:{solvinv}\:{sir} \\ $$

Commented by $@ty@m123 last updated on 09/Sep/19

You′re welcome. Meanwhile trying for (1)

$${You}'{re}\:{welcome}. \\ $$$${Meanwhile}\:{trying}\:{for}\:\left(\mathrm{1}\right) \\ $$

Answered by $@ty@m123 last updated on 10/Sep/19

(1) (a+c)cos (B/4)−2ccos (B/2)cos (B/4) =R(sin A+sin C)cos (B/4)−2Rsin Ccos (B/2)cos (B/4) =R.2sin ((A+C)/2)cos ((A−C)/2)cos (B/4)−2Rsin Ccos (B/2)cos (B/4) =R.2sin ((π−B)/2)cos ((A−C)/2)cos (B/4)−2Rsin Ccos (B/2)cos (B/4) =R.2cos (B/2)cos ((A−C)/2)cos (B/4)−2Rsin Ccos (B/2)cos (B/4) =2Rcos (B/2)cos (B/4){cos ((A−C)/2)−sin C} =2Rcos (B/2)cos (B/4){cos ((A−C)/2)−cos ((π−2C)/2)} =2Rcos (B/2)cos (B/4).2sin ((A−C+π−2C)/2)sin ((π−2C−A+C)/2) =2Rcos (B/2)cos (B/4).2sin ((π/2)−((A−3C)/2))sin ((π/2)−((A+C)/2)) =2Rcos (B/2)cos (B/4).2cos (((A−3C)/2))cos (((A+C)/2)) =2Rcos (B/2)cos (B/4).2cos (((A−3C)/2))cos (((π−B)/2)) =2Rcos (B/2)cos (B/4).2cos (((A−3C)/2))sin (B/2) =Rsin Bcos (B/4).2cos (((A−3C)/2)) =b.2cos (B/4).cos (((A−3C)/2)) =b.2cos (B/4).cos (((π−(B+C)−3C)/2)) =b.2cos (B/4).cos (((π−B−4C)/2)) =b.2cos (B/4).sin (((B+4C)/2)) Got stuck here. Either question or my path is wrong.

$$\left(\mathrm{1}\right)\:\left({a}+{c}\right)\mathrm{cos}\:\frac{{B}}{\mathrm{4}}−\mathrm{2}{c}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}} \\ $$$$={R}\left(\mathrm{sin}\:{A}+\mathrm{sin}\:{C}\right)\mathrm{cos}\:\frac{{B}}{\mathrm{4}}−\mathrm{2}{R}\mathrm{sin}\:{C}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}} \\ $$$$={R}.\mathrm{2sin}\:\frac{{A}+{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}−\mathrm{2}{R}\mathrm{sin}\:{C}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}} \\ $$$$={R}.\mathrm{2sin}\:\frac{\pi−{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}−\mathrm{2}{R}\mathrm{sin}\:{C}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}} \\ $$$$={R}.\mathrm{2cos}\:\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}−\mathrm{2}{R}\mathrm{sin}\:{C}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}} \\ $$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}\left\{\mathrm{cos}\:\frac{{A}−{C}}{\mathrm{2}}−\mathrm{sin}\:{C}\right\} \\ $$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}\left\{\mathrm{cos}\:\frac{{A}−{C}}{\mathrm{2}}−\mathrm{cos}\:\frac{\pi−\mathrm{2}{C}}{\mathrm{2}}\right\} \\ $$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2sin}\:\frac{{A}−{C}+\pi−\mathrm{2}{C}}{\mathrm{2}}\mathrm{sin}\:\frac{\pi−\mathrm{2}{C}−{A}+{C}}{\mathrm{2}} \\ $$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{{A}+{C}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2cos}\:\left(\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right)\mathrm{cos}\:\:\left(\frac{{A}+{C}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2cos}\:\left(\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right)\mathrm{cos}\:\:\left(\frac{\pi−{B}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}{R}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2cos}\:\left(\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right)\mathrm{sin}\:\:\frac{{B}}{\mathrm{2}} \\ $$$$={R}\mathrm{sin}\:{B}\mathrm{cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{2cos}\:\left(\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right) \\ $$$$={b}.\mathrm{2cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{cos}\:\left(\frac{{A}−\mathrm{3}{C}}{\mathrm{2}}\right) \\ $$$$={b}.\mathrm{2cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{cos}\:\left(\frac{\pi−\left({B}+{C}\right)−\mathrm{3}{C}}{\mathrm{2}}\right) \\ $$$$={b}.\mathrm{2cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{cos}\:\left(\frac{\pi−{B}−\mathrm{4}{C}}{\mathrm{2}}\right) \\ $$$$={b}.\mathrm{2cos}\:\frac{{B}}{\mathrm{4}}.\mathrm{sin}\:\:\left(\frac{{B}+\mathrm{4}{C}}{\mathrm{2}}\right) \\ $$$${Got}\:{stuck}\:{here}. \\ $$$${Either}\:{question}\:{or}\:{my}\:{path}\:{is}\:{wrong}. \\ $$

Commented by peter frank last updated on 14/Sep/19

$${thank}\:{you} \\ $$$$ \\ $$

Commented by peter frank last updated on 14/Sep/19

$${thank}\:{you} \\ $$$$ \\ $$

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