Question

If $E=co{s}^{2}71^o+co{s}^{2}49^o+cos71^{o}cos49^o,$ then the value of $10E$ is equal to

• A. $7.5$
• B. $2.5$
• C. $12.5$
• D. $10.5$

Answer 1

Answer: (A)

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