Sin A-sin B-3 cos B-cos A -sin 3 A-sin 3 B-

The correct option is A 0Sin A + sin B = √(3) (cos B cos A) ⇒√(3) cos A+sin A=√(3) cos B sin B ⇒ cos A ( √(3)/2 )+ sin A ( 1/2 )=cos B ( √(3)/2 ) sin B ( 1/2 ...

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Standard XI

Mathematics

Basic Trigonometric Identities

sin A+sin B=√...

Question

sin A + sin B = $\sqrt{3}$ (cos B – cos A) $\Rightarrow$ sin 3A + sin 3B =

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Solution

$S i n A + s i n B = \sqrt{3} (c o s B - c o s A) \Rightarrow \sqrt{3} c o s A + s i n A = \sqrt{3} c o s B - s i n B$

$\Rightarrow c o s A (\frac{\sqrt{3}}{2}) + s i n A (\frac{1}{2}) = c o s B (\frac{\sqrt{3}}{2}) - s i n B (\frac{1}{2}) \Rightarrow c o s (A - \frac{π}{6}) = c o s (B + \frac{π}{6})$

$\Rightarrow A - \frac{π}{6} = \pm (B + \frac{π}{6}) \Rightarrow A = - B or A - B = \frac{π}{3}$

Clearly A = - B satisfies the given equation. $∴ 3 A = - 3 B \Rightarrow s i n 3 A = s i n (- 3 B) \Rightarrow s i n 3 A + s i n 3 B = 0$

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Similar questions

sin A + sin B = \sqrt{3} (cos B - cos A)

. Find the value of

sin 3 A + sin 3 B

sinA+sinB=√3(cosB-cosA) thensin3A+sin3B=?

Options 1)0

2)2

3)1

4) -1

Q. Let

sin A + sin B + sin C = 0

then the value of

sin 3 A + sin 3 B + sin 3 C

Q. If for a triangle

A B C

∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = 0

then

{sin}^{3} A + {sin}^{3} B + {sin}^{3} C =

\frac{{cos}^{3} A + {sin}^{3} A}{cos A + sin A} + \frac{{cos}^{3} A - {sin}^{3} A}{cos A - sin A}

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sin A + sin B = √3 (cos B – cos A) ⇒ sin 3A + sin 3B =

SinA+sinB=√3 (cos B− cos A)⇒√3 cos A+sin A=√3 cos B−sin B ⇒cos A(√32)+sin A(12)=cos B(√32)−sin B(12)⇒cos(A−π6)=cos(B+π6) ⇒A−π6=±(B+π6)⇒A = - B or A - B =π3 Clearly A = - B satisfies the given equation. ∴3A=−3B ⇒sin 3A=sin(−3B)⇒sin 3A+sin 3B=0

sin A + sin B = $\sqrt{3}$ (cos B – cos A) $\Rightarrow$ sin 3A + sin 3B =