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Transformation Formulae — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae2 Marks
Question
Prove that:
$\cos80^\circ+\cos40^\circ-\cos20^\circ=0$
✓

Answer

$\text{LHS}=\cos80^\circ+\cos40^\circ-\cos20^\circ$
$(\cos80^\circ+\cos40^\circ)-\cos20^\circ$
$=\ 2\cos\Big(\frac{80^\circ+40^\circ}{2}\Big)\cos\Big(\frac{80^\circ-40^\circ}{2}\Big)-\cos20^\circ$ $\Big[\because\ \sin\text{c}-\sin\text{D}=2\sin\Big(\frac{\text{C}-\text{D}}{2}\Big)\cos\Big(\frac{\text{C+D}}{2}\Big)\Big]$
$=\ 2\cos60^\circ\cos20^\circ-\cos20^\circ$
$=\ 2\times\frac{1}{2}\cos20-\cos20^\circ$
$=\ \cos20^\circ-\cos20^\circ$
$=\ 0$
$=\ \text{RHS}$

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