Let A - B - C are three angles such thatsin A-sin B-sin C-0- then the value of sin A -sin B -sin C-sin 3 A-sin 3 B-sin 3 C wherever defined isA- -12B- 12C- -1-12D- 1-12

The correct option is C 1/12sin A.sin B .sin C/sin 3A+sin 3B +sin 3C =sin A.sin B .sin C/3(sin A+sin B +sin C) 4(sin^3 A+sin^3 B +sin^3 C) =sin A.sin B .sin C ...

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Standard XII

Mathematics

Solving Simultaneous Trigonometric Equations

Let A , B , C...

Question

Let $A, B, C$ are three angles such that
$sin A + sin B + sin C = 0,$ then the value of $\frac{sin A . sin B . sin C}{sin 3 A + sin 3 B + sin 3 C}$ (wherever defined) is

12

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- \frac{1}{12}

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\frac{1}{12}

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Solution

The correct option is C $- \frac{1}{12}$
$\frac{sin A . sin B . sin C}{sin 3 A + sin 3 B + sin 3 C} = \frac{sin A . sin B . sin C}{3 (sin A + sin B + sin C) - 4 ({sin}^{3} A + {sin}^{3} B + {sin}^{3} C)} = \frac{sin A . sin B . sin C}{- 4 (3 sin A . sin B . sin C)} = - \frac{1}{12}$

$[a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) \Rightarrow a^{3} + b^{3} + c^{3} = 3 a b c if a + b + c = 0]$

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Let A,B,C are three angles such that sinA+sinB+sinC=0, then the value of sinA.sinB.sinCsin3A+sin3B+sin3C (wherever defined) is

The correct option is C −112sinA.sinB.sinCsin3A+sin3B+sin3C=sinA.sinB.sinC3(sinA+sinB+sinC)−4(sin3A+sin3B+sin3C)=sinA.sinB.sinC−4(3sinA.sinB.sinC)=−112 [ a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)⇒a3+b3+c3=3abc if a+b+c=0]

Let $A, B, C$ are three angles such that
$sin A + sin B + sin C = 0,$ then the value of $\frac{sin A . sin B . sin C}{sin 3 A + sin 3 B + sin 3 C}$ (wherever defined) is