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Transformation Formulae — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae2 Marks
Question
Prove that:
$\sin50^\circ-\sin70^\circ+\sin10^\circ=0$
✓

Answer

$\text{LHS}=\sin50^\circ-\sin70^\circ+\sin10^\circ$
$(\sin50^\circ-\sin70^\circ)+\sin10^\circ$
$=\ \Big(2\sin\Big(\frac{50^\circ-70^\circ}{2}\Big)\cos\Big(\frac{50^\circ+70^\circ}{2}\Big)\Big)+\sin10^\circ$ $\Big[\because\ \sin\text{c}-\sin\text{D}=2\sin\Big(\frac{\text{C}-\text{D}}{2}\Big)\cos\Big(\frac{\text{C+D}}{2}\Big)\Big]$
$=\ 2\sin(-10^\circ)\cos60^\circ+\sin10^\circ$
$=\ -2\sin10^\circ\times\frac{1}{2}+\sin10^\circ$ $\Big[\because\ \cos60^\circ=\frac{1}{2}\Big]$
$=\ 0$
$=\ \text{RHS}$

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