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  4. If E=cos2 71o+cos2⁡49o+cos⁡71ocos⁡49o, then the value of 10E is equal to

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Q. If $E=cos^{2} 71^{o}+cos^{2}⁡49^{o}+cos⁡71^{o}cos⁡49^{o},$ then the value of $10E$ is equal to

NTA AbhyasNTA Abhyas 2020

Solution:

$E=\frac{1 + cos \left(142\right)^{o}}{2}+\frac{1 + cos ⁡ \left(98\right)^{o}}{2}+\frac{1}{2}\left(2 cos ⁡ \left(71\right)^{o} cos ⁡ \left(49\right)^{o}\right)$
$=1+\frac{1}{2}\left(cos \left(142\right)^{o} + cos ⁡ \left(98\right)^{o}\right)+\frac{1}{2}\left(cos ⁡ \left(120\right)^{o} + cos ⁡ \left(22\right)^{o}\right)$
$=1+\frac{1}{2}\left(2 cos \left(120\right)^{o} cos ⁡ \left(22\right)^{o}\right)+\frac{1}{2}\left(- \frac{1}{2} + cos ⁡ \left(22\right)^{o}\right)$
$=1+\left(- \frac{1}{2}\right)cos \left(22\right)^{o}-\frac{1}{4}+\frac{1}{2}cos ⁡ \left(22\right)^{o}=\frac{3}{4}$
$E=\frac{3}{4}=0.75$
$\Rightarrow 10E=7.5$