If A - B - C - 180- then sin2A2 - sin2B2 - sin2C2 -

The correct option is D 1 2 sinA2sinB2sinC2 Let sin^2A2 + sin^2B2 + sin^2C2 =I I=sin^2A2 + (1 cos^2B2 ) + sin^2C2 =sin^2A2 cos^2B2+sin^2 ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If A + B + ...

Question

If $A + B + C = 180^{\circ}$ , then ${sin}^{2} \frac{A}{2} + {sin}^{2} \frac{B}{2} + {sin}^{2} \frac{C}{2} =$

1 + 2 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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1 + 2 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

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1 - 2 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

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1 - 2 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

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Solution

The correct option is D $1 - 2 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}$
Let ${sin}^{2} \frac{A}{2} + {sin}^{2} \frac{B}{2} + {sin}^{2} \frac{C}{2} = I$
$I = s i n^{2} \frac{A}{2} + (1 - c o s^{2} \frac{B}{2}) + s i n^{2} \frac{C}{2}$
$= {sin}^{2} \frac{A}{2} - {cos}^{2} \frac{B}{2} + {sin}^{2} \frac{C}{2} + 1$
$= - cos (\frac{A + B}{2}) cos (\frac{A - B}{2}) + {sin}^{2} \frac{C}{2} + 1$
As $A + B + C = 180^{\circ} \Rightarrow \frac{A + B}{2} = 90^{\circ} - \frac{C}{2}$
$I = - cos (90^{\circ} - \frac{C}{2}) cos (\frac{A - B}{2}) + {sin}^{2} \frac{C}{2} + 1$
$= - sin (\frac{C}{2}) cos (\frac{A - B}{2}) + {sin}^{2} \frac{C}{2} + 1$
$= - sin (\frac{C}{2}) (cos (\frac{A - B}{2}) - sin (\frac{C}{2})) + 1$
Now using $\frac{C}{2} = 90^{\circ} - \frac{A + B}{2}$
$I = - sin (\frac{C}{2}) (cos (\frac{A - B}{2}) - sin (90^{o} - \frac{A + B}{2})) + 1$
$= - sin (\frac{C}{2}) (c o s (\frac{A - B}{2}) - c o s (\frac{A + B}{2})) + 1$
$= - 2 sin (\frac{A}{2}) sin (\frac{B}{2}) sin (\frac{C}{2}) + 1$
Hence, option $^{'} C^{'}$ is correct.

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