Question

Evaluate: $co{s}^2{10}^0+co{s}^2{50}^0+co{s}^2{70}^0$

Answer 1

$co{s}^{2}10+co{s}^{2}50+co{s}^{2}70$

$=\frac{3+cos20+cos100+cos140}{2}$

Formula: $co{s}^2\theta =\frac{1+cos2\theta }{2}$

$=\frac{3+cos20-(sin10+sin50)}{2}$

Since $cos100=sin10$ and $cos140=-sin50$

formula: $sinA+sinB=2sin\frac{A+B}{2}\cdot cos\frac{A-B}{2}$

$=\frac{3+cos20-2sin30\cdot cos20}{2}$

$=\frac{3+cos20-2\cdot \frac{1}{2}cos20}{2}$

$=\frac{3+cos20-cos20}{2}$

$=\frac{3}{2}$