Prove that- sin2A2-sin2B2-sin2C2-1-2cosA2cosB2sinC2-

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Standard XII

Mathematics

Basic Trigonometric Identities

Prove that: ...

Question

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

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Solution

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

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

We want +1 in R.H.S.L.H.S. =1−cos2(A/2)+sin2(B/2)−sin2(C/2)=1−cos2(A/2)+[sin{(B+C)/2}sin{(B−C)/2}]=1−cos(A/2)[cosA/2−{sin(B−C)/2}]=1−cos(A/2)[sin{(B+C)/2−sin(B−C)/2}]=1−cos(A/2)[2cos(B/2)sin(C/2)]=1−2cos(A/2)cos(B/2)sin(C/2).

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