Question

Find the greatest and least value of $y$,
$y=10{\cos }^{2}x-6\sin x\cos x+2{\sin }^{2}x$

Answer 1

$y=10{\cos }^{2}x-6\sin x\cos x+2{\sin }^{2}x$

$Greatest$

$=11\left({\cos }^{2}x+{\sin }^{2}x\right)-\left({\cos }^{2}x+6\sin x\cos x+9{\sin }^{2}x\right)$

$=11-{\left(\cos x+3\sin x\right)}^2\Rightarrow y\le 11$

$Least=\left({\cos }^{2}x+{\sin }^{2}x\right)-\left(9{\cos }^{2}x-6\sin x\cos x+{\sin }^{2}x\right)$

$=1+{\left(3\cos x-\sin x\right)}^2\ge 1$

$\Rightarrow So,\left[1,11\right]$