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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ=\frac{3}{16}$
✓

Answer

$\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ=\frac{3}{16}$
$\text{LHS}=\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ$
$=\ \cos30^\circ\cos10^\circ\cos50^\circ\cos70^\circ$
$=\ \frac{\sqrt3}{2}(\cos10^\circ\cos50^\circ\cos70^\circ)$
$=\ \frac{\sqrt3}{2}(\cos10^\circ\cos50^\circ)\cos70^\circ$
$=\ \frac{\sqrt3}{2}(2\cos10^\circ\cos50^\circ)\cos70^\circ$$[\text{Multiplying and dividing by 2}]$
Also,
$\Rightarrow\ 2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})\ \dots(\text{i})$
$=\ \frac{\sqrt3}{4}\cos70^\circ(\cos(50^\circ+10^\circ)+\cos(10^\circ-50^\circ))$
$=\ \frac{\sqrt3}{4}\cos70^\circ(\cos60^\circ+(-40^\circ))$
Now,
$\cos(-\theta)=\cos\theta$
$=\ \frac{\sqrt3}{4}\cos70^\circ\Big(\frac{1}{2}+\cos40^\circ\Big)\Big[\because\cos60^\circ=\frac{1}{2}\Big]$
$=\ \frac{\sqrt3}{8}\cos70^\circ+\frac{\sqrt3}{4}\cos70^\circ\cos40^\circ$
$=\ \frac{\sqrt3}{8}\cos70^\circ+\frac{\sqrt3}{8}(2\cos70^\circ\cos40^\circ)$
$=\ \frac{\sqrt3}{8}[\cos70^\circ+\cos(70^\circ+40^\circ)+\cos(70^\circ-40^\circ)][\text{from(i)}]$
$=\ \frac{\sqrt3}{8}[\cos70^\circ+\cos110^\circ+\cos30^\circ]$
$=\ \frac{\sqrt3}{8}\Big[\cos70^\circ+\cos(180^\circ-70^\circ)+\frac{\sqrt3}{2}\Big]$
$=\ \frac{\sqrt3}{8}\Big[\cos70^\circ-\cos70^\circ+\frac{\sqrt3}{2}\Big][\because\cos(180^\circ-\theta)=-\cos\theta]$
$=\ \frac{\sqrt3}{8}\times\frac{\sqrt3}{2}=\frac{3}{16}$
$=\ \text{RHS}$

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