Question

Cos 6° • Cos 42° • Cos 66° • Cos78° = $\frac{1}{16}$​

Answer 1

Formula Applied :

$\sf \bigstar\ \; 2.cosx.cosy=cos(x+y)cos(x-y)$

$\bigstar\ \; \sf cos(120)=cos(90+30)=-sin30=-\dfrac{1}{2}$

$\bigstar\ \; \sf cosx-cosy=-2.sin \left(\dfrac{x+y}{2} \right) .sin\left(\dfrac{x-y}{2}\right)$

$\bigstar\ \; \sf sin(90-\theta)=cos\theta$

$\bigstar\ \; \sf sin2\theta=2.sin\theta.cos\theta$

$\bigstar\ \; \sf 2.sinx.siny=cos(x-y)-cos(x+y)$

Solution :

LHS

$\to \sf cos\ 6.cos\ 66.cos\ 42.cos\ 78$

$\to \sf \dfrac{1}{4}(2.cos\ 6.cos\ 66)(2.cos\ 42.cos\ 78)$

Apply 1st formula ,

$\\ \to \sf \dfrac{1}{4}\left\{cos(66+6)+cos(66-6) \right\}\left\{cos(78+42)+cos(78-42)\right\}$

$\\ \to \sf \dfrac{1}{4}\left( cos(72)+cos(60)\right)\left( cos(120)+cos(36)\right)$

$\\ \to \sf \dfrac{1}{4}\left(cos\ 72+\dfrac{1}{2}\right)\left(-\dfrac{1}{2}+cos\ 36\right)$

$\\ \to \sf \dfrac{1}{4}\left(-\dfrac{1}{2}cos\ 72+cos\ 36.cos\ 72-\dfrac{1}{2}.\dfrac{1}{2}+\dfrac{1}{2}cos\ 36\right)$

$\\ \to \sf \dfrac{1}{4}\left(- \dfrac{1}{4}+cos\ 36.cos\ 72+\dfrac{1}{2}(cos\ 36-cos\ 72)\right)$

Apply 3rd formula ,

$\\ \to \sf \dfrac{1}{4}\left( -\dfrac{1}{4}+cos\ 36.cos\ 72+\dfrac{1}{2}\left(-2sin\dfrac{36+72}{2}.sin\dfrac{36-72}{2} \right) \right)$

$\\ \to \sf \dfrac{1}{4}\left( -\dfrac{1}{4}+cos\ 36.cos\ 72+\dfrac{1}{2}(-2sin\ 54.(-sin\ 18))\right)$

$\\ \to \sf \dfrac{1}{4}\left(-\dfrac{1}{4}+cos\ 36.cos\ 72+sin\ 18.sin\ 54\right)$

$\\ \to \sf \dfrac{1}{4}\left( -\dfrac{1}{4}+sin(90-36).sin(90-72)+sin\ 18.sin\ 54 \right)$

$\\ \to \sf \dfrac{1}{4}\left(-\dfrac{1}{4}+sin\ 54.sin\ 18+sin\ 54.sin\ 18 \right)$

$\\ \to \sf \dfrac{1}{4}\left(-\dfrac{1}{4}+2sin\ 54.sin\ 18 \right)$

$\\ \sf \to \sf \dfrac{1}{4}\left(-\dfrac{1}{4}+2sin\ 54.sin\ 18\times \dfrac{cos\ 18}{cos\ 18} \right)$

Apply formula 5 ,

$\\ \to \sf \dfrac{1}{4}\left(-\dfrac{1}{4}+\dfrac{2.sin\ 54.sin\ 36}{2.cos\ 18}\right)$

Apply formula 6 ,

$\\ \to \sf \dfrac{1}{4}\left( -\dfrac{1}{4}+\dfrac{cos\ 18-cos\ 90}{2\ cos\ 18}\right)$

$\\ \to \sf \dfrac{1}{4}\left(-\dfrac{1}{4}+\dfrac{1}{2}\right)$

$\\ \to \sf \dfrac{1}{4}\left( \dfrac{1}{4}\right)$

$\\ \to \dfrac{1}{16}$

RHS