- Home
- Standard 11
- Mathematics
3.Trigonometrical Ratios, Functions and Identities
easy
ત્રિકોણ $ABC$ માટે , $\sin 2A + \sin 2B + \sin 2C = . . ..$
A
$4\sin A.\,\sin B.\,\sin C$
B
$4\cos A.\,\cos B.\,\cos C$
C
$2\cos A.\,\cos B.\,\cos C$
D
$2\sin A.\,\sin B.\,\,\sin C$
Solution
(a) We know that $A + B + C = 180^\circ $ (in $\Delta ABC$)
Now, $\sin 2A + \sin 2B + \sin 2C$
$ = 2\sin (A + B)\cos (A – B) + 2\sin C\cos C$
$ = 2\sin (\pi – C)\cos (A – B) + 2\sin C\cos (\pi – \overline {A + B} )$
$ = 2\sin C\cos (A – B) – 2\sin C\cos (A + B)$
$ = 2\sin C\{ \cos (A – B) – \cos (A + B)\} $
$ = 2\sin C\{ 2\sin A\sin B\} = 4\sin A\sin B\sin C$.
Standard 11
Mathematics
