If A+B+C=90° then prove that ${sin}^{2} (\frac{A}{2}) + {sin}^{2} (\frac{B}{2}) + {sin}^{2} (\frac{C}{2}) = 1 - 2 sin A . sin B . sin C$ $sin^2 (A/2)+sin^2 (B/2) + sin^2 (C/2) = 1-2sinA.sinB.sinC$ ?

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If A+B+C=90° then prove that ${sin}^{2} (\frac{A}{2}) + {sin}^{2} (\frac{B}{2}) + {sin}^{2} (\frac{C}{2}) = 1 - 2 sin A . sin B . sin C$ $sin^2 (A/2)+sin^2 (B/2) + sin^2 (C/2) = 1-2sinA.sinB.sinC$ ?

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Answer 1 · 2018-04-29T14:19:58

Fun. Let's check it before we spend too much time on it.

For the easiest numbers, let $A = 90^{\circ}, B = C = 0^{\circ} .$ $A=90^circ, B=C=0^circ.$ We get ${sin}^{2} 45^{\circ} = \frac{1}{2}$ $sin^2 45^circ = 1/2$ on the left and $1 - 2 \ sin 90^circ sin 0 sin 0 = 1$ on the right. It's false. Cue the deflated trombone, wah wah waaah.

Answer 2 · 2018-04-29T14:26:22

question is incorrect.

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If A+B+C=90° then prove that ${sin}^{2} (\frac{A}{2}) + {sin}^{2} (\frac{B}{2}) + {sin}^{2} (\frac{C}{2}) = 1 - 2 sin A . sin B . sin C$ $sin^2 (A/2)+sin^2 (B/2) + sin^2 (C/2) = 1-2sinA.sinB.sinC$ ?

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If A+B+C=90° then prove that sin^2 (A/2)+sin^2 (B/2) + sin^2 (C/2) = 1-2sinA.sinB.sinCsin2(A2)+sin2(B2)+sin2(C2)=1−2sinA.sinB.sinCsin^2 (A/2)+sin^2 (B/2) + sin^2 (C/2) = 1-2sinA.sinB.sinC ?

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If A+B+C=90° then prove that ${sin}^{2} (\frac{A}{2}) + {sin}^{2} (\frac{B}{2}) + {sin}^{2} (\frac{C}{2}) = 1 - 2 sin A . sin B . sin C$ $sin^2 (A/2)+sin^2 (B/2) + sin^2 (C/2) = 1-2sinA.sinB.sinC$ ?