# Question 8. . . . . . . . . . . . . . ?? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Ajay
209 Points
8 years ago
$\dpi{100} Given\quad A+B+C\quad =\quad 180-----(1)\\ { sin }^{ 2 }A+{ sin }^{ 2 }B+{ sin }^{ 2 }C\\ =1/2(1-cos2A+1-cos2B+1-cos2C)\\ =1/2\left\{ 3-\left( cos2A+cos2B+cos2C \right) \right\} \\ =1/2\left\{ 3-\left( 2cos(A+B)cos(A-B)+cos2C \right) \right\} \\ From(1)\quad cos2C\quad =\quad cos(360-2A-2B)\quad \\ =\quad cos2(A+B)\quad =\quad 2{ cos }^{ 2 }(A+B)-1\\ Substituting\quad cos2C\quad we\quad have\\ =\quad 1/2\left\{ 3-\left( 2cos(A+B)cos(A-B)+2{ cos }^{ 2 }(A+B)-1 \right) \right\} \\ =1/2\left\{ 4-2cos(A+B)\left( cos(A-B)+{ cos }(A+B) \right) \right\} \\ from\quad (1)\quad cos(A+B)\quad =\quad cosC\quad \\ \\ =1/2(4-4cosCcosAcosB)\\ =2(1-cosCcosAcosB)\\$
Ajay
209 Points
8 years ago
Slight correction last few steps....................................................................................................
$\dpi{100} from\quad (1)\quad cos(A+B)\quad =\quad -cosC\quad \\ \\ =1/2(4+4cosCcosAcosB)\\ =2(1+cosCcosAcosB)$