Question

${\cos }^2({60}^0-x)+{\cos }^2({60}^0+x)\in$

• A. $[-\frac{1}{2},\frac{1}{2}]$
• B. $[\frac{1}{2},1]$
• C. $[\frac{1}{2}\frac{3}{2}]$
• D. $[\frac{3}{2},2]$

Answer 1

The correct option is C $[\frac{1}{2}\frac{3}{2}]$

$f\left(x\right)=co{s}^2(60-x)+co{s}^2(60+x)$
$=\left[cos\right(60-x)+cos(60+x){]}^2-2cos(60-x\left)cos\right(60+x)$
$=\left(2cos\right({60}^o)cosx{)}^2-[cos2x+cos120]$
$=co{s}^{2}x-[-\frac{1}{2}+cos2x]$
$=co{s}^{2}x-(2co{s}^{2}x-1)+\frac{1}{2}$
$f\left(x\right)=-co{s}^{2}x+\frac{3}{2}$
range of $f\left(x\right)ϵ[\frac{1}{2},\frac{3}{2}]$