Prove that- sin 2A-sin 2B-sin 2Csin A-sin B-sin C-8sin A2sin B2sin C2-

L.H.S. =4sin Asin Bsin C4cos (A/2)cos (B/2)cos (C/2) ={2sin (A/2)cos (A/2)}{2sin (B/2)cos (B/2)}{(2sin (C/2)cos (C/2)}cos (A/2)cos (B/2)cos (C/2) ...

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Standard XII

Mathematics

Basic Trigonometric Identities

Prove that: ...

Question

Prove that: $\frac{sin 2 A + sin 2 B + sin 2 C}{sin A + sin B + sin C} = 8 sin (\frac{A}{2}) sin (\frac{B}{2}) sin (\frac{C}{2})$ .

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Solution

L.H.S. $= \frac{4 sin A sin B sin C}{4 cos (A / 2) cos (B / 2) cos (C / 2)}$
$= \frac{{2 sin (A / 2) cos (A / 2)} {2 sin (B / 2) cos (B / 2)} {(2 sin (C / 2) cos (C / 2)}}{cos (A / 2) cos (B / 2) cos (C / 2)}$
$= 8 sin (\frac{A}{2}) sin (\frac{B}{2}) sin (\frac{C}{2})$ .

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Q. In a

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the expression

\frac{sin 2 A + sin 2 B + sin 2 C}{sin A + sin B + sin C} = k sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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Q. If

A + B + C = 180^{o}, \frac{sin 2 A + sin 2 B + sin 2 C}{sin A + sin B + sin C} = k sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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Prove that ${sin}^{2} (\frac{A}{2}) + {sin}^{2} (\frac{B}{2}) + {sin}^{2} (\frac{C}{2}) = 1 - 2 sin (\frac{A}{2}) sin \frac{B}{2}) sin (\frac{C}{2})$ .

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A + B + C = 180^{\circ}

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cos A + cos B + cos C

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If A+B+C=180degree. Prove that sin2A+sin2B+sin2C=4×sinA×sinB×sinC.

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Prove that: sin2A+sin2B+sin2CsinA+sinB+sinC=8sin(A2)sin(B2)sin(C2).

L.H.S. =4sinAsinBsinC4cos(A/2)cos(B/2)cos(C/2)={2sin(A/2)cos(A/2)}{2sin(B/2)cos(B/2)}{(2sin(C/2)cos(C/2)}cos(A/2)cos(B/2)cos(C/2)=8sin(A2)sin(B2)sin(C2).

Prove that: $\frac{sin 2 A + sin 2 B + sin 2 C}{sin A + sin B + sin C} = 8 sin (\frac{A}{2}) sin (\frac{B}{2}) sin (\frac{C}{2})$ .

L.H.S. $= \frac{4 sin A sin B sin C}{4 cos (A / 2) cos (B / 2) cos (C / 2)}$
$= \frac{{2 sin (A / 2) cos (A / 2)} {2 sin (B / 2) cos (B / 2)} {(2 sin (C / 2) cos (C / 2)}}{cos (A / 2) cos (B / 2) cos (C / 2)}$
$= 8 sin (\frac{A}{2}) sin (\frac{B}{2}) sin (\frac{C}{2})$ .