Prove that sin2A2-sin2B2-sin2C2-1-2sinA2sinB2sinC2-

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Standard XII

Mathematics

Basic Trigonometric Identities

Prove that ...

Question

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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Solution

We want $+ 1$ in R.H.S.
$∴$ L.H.S. $= 1 - {cos}^{2} (A / 2) + {sin}^{2} (B / 2) + {sin}^{2} (C / 2)$
$= 1 - {{cos}^{2} (A / 2) - {sin}^{2} (B / 2)} + {sin}^{2} (C / 2)$
$= 1 - cos {(A + B) / 2} cos {(A - B) / 2} + {sin}^{2} (C / 2)$
$= 1 - sin (C / 2) cos {(A - B) / 2} + {sin}^{2} (C / 2)$
$= 1 - sin (C / 2) [cos {(A - B) / 2} - sin (C / 2)]$
$= 1 - sin (C / 2) [cos {(A - B) / 2} - cos {(A + B) / 2}]$
$= 1 - sin (C / 2) [2 sin (A / 2) sin (B / 2)]$
$= 1 - 2 sin (A / 2) sin (B / 2) sin (C / 2)$ .

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Prove that sin2(A2)+sin2(B2)+sin2(C2)=1−2sin(A2)sinB2)sin(C2).

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