The minimum value of 3 cosx + 4 sinx + 9 is

1.	The minimum value of 3 cosx + 4 sinx + 9 is
Answer» $\implies$ ➩minimum VALUE of 3cosx + 4sinx + 8 = 3 ➩we have to find minimum value of 3cosx + 4sinx +8. ➩using application, ➩ -√(a² + b²) ≤ acosθ + bsinθ ≤ √(a² + b²) ➩where , a and B are rational numbers. then, -√(3² + 4²) ≤ 3cosx + 4sinx ≤ √(3² + 4²) ⇒-5 ≤ 3cosx + 4sinx ≤ 5 ➩minimum value of 3cosx + 4sinx = -5 ➩so, minimum value of 3cosx + 4sinx + 8 = -5 + 8 = 3

Answer»

$\implies$

➩minimum VALUE of 3cosx + 4sinx + 8 = 3

➩we have to find minimum value of 3cosx + 4sinx +8.

➩using application,

➩ -√(a² + b²) ≤ acosθ + bsinθ ≤ √(a² + b²)

➩where , a and B are rational numbers.

then, -√(3² + 4²) ≤ 3cosx + 4sinx ≤ √(3² + 4²)

⇒-5 ≤ 3cosx + 4sinx ≤ 5

➩minimum value of 3cosx + 4sinx = -5

➩so, minimum value of 3cosx + 4sinx + 8 = -5 + 8 = 3

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The minimum value of 3 cosx + 4 sinx + 9 is

➩using application,

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