Question

If a ≤ cos θ + 3√2 sin ($\theta + \frac{\pi}{4}$) + 6 ≤ b, then find the largest value of a and smallest value of b.

Answer 1

it is given that a ≤ cosθ + 3√2sin(θ + π/4) + 6 ≤ b
we have to find largest value of a and smallest value of b.

we know, -√(r² + s²) ≤ rcosx + s sinx ≤ √(r² + s²)
[ this can be proved with help of trigonometric concepts , you just remember this application ]

so, here, cosθ + 3√2sin(θ + π/4)

= cosθ + 3√2[sinθ.cosπ/4 + sinπ/4 cosθ]

= cosθ + 3√2[ 1/√2 sinθ + 1/√2 cosθ]

= cosθ + 3sinθ + 3cosθ

= 4cosθ + 3sinθ .......(1)

here, r = 4 and s = 3

so, -√(4² + 3²) ≤ 4cosθ + 3sinθ ≤ √(4² + 3²)

or, -5 ≤ 4cosθ + 3sinθ ≤ 5

or, -5 ≤ cosθ + cos(θ + π/4) ≤ 5 [ from equation (1), ]

or, -5 + 6 ≤ cosθ + cos(θ + π/4) + 6 ≤ 5 + 6

or, 1 ≤ cosθ + cos(θ + π/4) + 6 ≤ 11

compare with a ≤ cosθ + 3√2sin(θ + π/4) + 6 ≤ b
we get, a = 1 and b = 11

hence, a = 1 and b = 11