Question

find minimum and maximum values of 9 cos^2 x + 48 sin x cos x - 5 sin^2 x - 2​

Answer 1

9cos²x + 48sinx.cosx - 5sin²x - 2

use formula,

cos²x = (1 + cos2x)/2

sin²x = (1 - cos2x)/2

sinx.cosx = sin2x/2

so, 9(1 + cos2x)/2 + 24sin2x - 5(1- cos2x)/2 - 2

= 9/2 + 9/2 cos2x + 24sin2x - 5/2 + 5/2 cos2x - 2

= (9 - 5)/2 + (9 + 5)/2 cos2x + 24sin2x - 2

= 2 + 7cos2x + 24sin2x - 2

= 24sin2x + 7cos2x

we know, -√(a² + b²) ≤ asinθ + bcosθ ≤ √(a² + b²)

so, -√(24² + 7²) ≤ 24sin2x + 7cos2x ≤ √(24² + 7²)

⇒- 25 ≤ 24sin2x + 7cos2x ≤ 25

hence, maximum value of expression is 25 and minimum value is -25.

Answer 2

$\huge{\pink {\sf\underline{Step\: by\: step\: explanation}}}$

9cos²x + 48sinx.cosx - 5sin²x - 2

use formula,

cos²x =${(1 + cos2x)}{2}$ <br>

sin²x =$\dfrac{(1 - cos2x)}{2}$

sinx.cosx = $\dfrac{sin2x}{2}$

so, $\dfrac{9(1 + cos2x)}{2} + 24sin2x - \dfrac{5(1- cos2x)}{2 - 2}$

$= \dfrac{9}{2} + \dfrac{9}{2} cos2x + 24sin2x -\dfrac {5}{2} +\dfrac {5}{2} cos2x - 2$

$= \dfrac{(9 - 5)}{2} +\dfrac {(9 + 5)}{2} cos2x + 24sin2x - 2$

= 2 + 7cos2x + 24sin2x - 2

= 24sin2x + 7cos2x

we know, -√(a² + b²) ≤ asinθ + bcosθ ≤ √(a² + b²)

so, -√(24² + 7²) ≤ 24sin2x + 7cos2x ≤ √(24² + 7²)

⇒- 25 ≤ 24sin2x + 7cos2x ≤ 25

So, maximum value of expression is 25 and minimum value is -25.