Question

If $si{n}^2$ x + $co{s}^2$y = 2 $se{c}^2$ z, find the value of $co{s}^2$ x + $si{n}^2$ y + 2 $si{n}^2$ z.

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Answer 1

Maximum value of $si{n}^2$x and $co{s}^2$y is one. Minimum value of $se{c}^2$z is one and minimum value of 2$se{c}^2$z is two.

So we have.

$si{n}^2$x + $co{s}^2$y ≤ 2 and $si{n}^2$x + $co{s}^2$y = 2 $se{c}^2$z ≥ 2

$\Rightarrow$ $si{n}^2$x + $co{s}^2$y = 2
This means each of the values $si{n}^2$x, $co{s}^2$y and $se{c}^2$z is equal to one.

$si{n}^2$x = 1 $\Rightarrow$ $co{s}^2$x = 0

$co{s}^2$y = 1 $\Rightarrow$ $si{n}^2$ y = 0

$se{c}^2$z = 1 $\Rightarrow$ $co{s}^2$z = 1 $\Rightarrow$ $si{n}^2$z = 0
$\Rightarrow$ $co{s}^2$x + $si{n}^2$ y +2 $si{n}^2$ z=0+0+ 2 X 0

=0