Question

Find the minimum value of 10 $co{s}^{2}x$ - 6 sinxcos x + 2 $si{n}^{2}x$

__

Answer 1

In this case also,we will try to convert the expression in the form a cos$\theta$ + bsin$\theta$.

We see sinxcosx, $co{s}^{2}x$ and $si{n}^{2}x$ in the expression.So we will simplify the terms with sin2x and cos2x

10$co{s}^{2}x$ - 6sinx cosx + 2$si{n}^{2}x$

= 8 $co{s}^{2}x$ - 3 sin2x + 2

= 4 (1 + cos2x) - 3 sin2x + 2

= 6 + 4 cos2x - 3sin2x

$\Rightarrow$ Maximum value = 6 + maximum value of (4 cos2x - 3 sin2x)

= 6 + 5

= 11

Minimum value = 6 + minimum value of (4 cos2x - 3 sin2x)

= 6 + (-5)

= 1

Key steps : (1) Expressing the given expression in terms of sin2x and cos2x