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Trigonometrical Identities — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsTrigonometrical Identities2 Marks
Question
For triangle $\text{ABC},$ show that : $\frac{\sin (A+B)}{2}=\frac{\cos C}{2}$
✓

Answer

We know that for a triangle $\triangle ABC$
$\angle A +\angle B +\angle C = 180^\circ$
$\frac{\angle B+\angle A}{2}$
$=90-\frac{\angle C}{2}$
$\sin \left(\frac{A+B}{2}\right)$
$=\sin \left(90^{\circ}-\frac{C}{2}\right)$
$=\cos \left(\frac{C}{2}\right)$

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