Question

Minimum value of 3cosX+4sinX+8

Answer 1

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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Answer 2

The minimum value of this question is 3