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Express 3A+ 3B+ 3C as the product of three trigonometrical ra

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Express $sin 3 A + sin 3 B + sin 3 C$ as the product of three trigonometrical ratios where A, B, C are the angles of a triangle. If the given expression be zero, then at least one angle of the triangle is $6 0^{o}$

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Eleanor Discussed

Express

sin 3 A + sin 3 B + sin 3 C

as the product of three trigonometrical ratios where A, B, C are the angles of a triangle. If the given expression be zero, then at least one angle of the triangle is

6 0^{o}

8 mins ago

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Text solutionVerified

A + B + C = π, ∴ 3 A + 3 B + 3 C = 3 π

\frac{1}{2} (3 A + 3 B) = \frac{3}{2} π - \frac{3}{2} C

∴ sin \frac{1}{2} (3 A + 3 B) = - cos \frac{3}{2} C

and

cos \frac{1}{2} (3 A + 3 B) = - sin \frac{3}{2} C

.
L.H.S.

= 2 sin \frac{3 A + 3 B}{2} cos \frac{3 A - 3 B}{2} + 2 sin \frac{3 C}{2} cos \frac{3 C}{2}

= - 2 cos \frac{3 C}{2} cos \frac{3 A - 3 B}{2} + 2 sin \frac{3 C}{2} cos \frac{3 C}{2}

= - 2 cos \frac{3 C}{2} [cos \frac{3 A - 3 B}{2} - sin \frac{3 C}{2}]

= - 2 cos \frac{3 C}{2} [cos \frac{3 A - 3 B}{2} + cos \frac{3 A + 3 B}{2}]

= - 4 cos \frac{3 A}{2} cos \frac{3 B}{2} cos \frac{3 C}{2}

If it be zero then at least one factor is zero, hence

3 A /2 = 9 0^{o}

A = 6 0^{o}

etc.

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Question Text	Express $sin 3 A + sin 3 B + sin 3 C$ as the product of three trigonometrical ratios where A, B, C are the angles of a triangle. If the given expression be zero, then at least one angle of the triangle is $6 0^{o}$
Answer Type	Text solution:1
Upvotes	150

Express sin3A+sin3B+sin3C as the product of three trigonometrical ratios where A, B, C are the angles of a triangle. If the given expression be zero, then at least one angle of the triangle is 60o

Text solutionVerified

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Express $sin 3 A + sin 3 B + sin 3 C$ as the product of three trigonometrical ratios where A, B, C are the angles of a triangle. If the given expression be zero, then at least one angle of the triangle is $6 0^{o}$