Prove that (2√3 + 3) sin x + 2√3 cos x lies between – (2√3 +

1.	Prove that (2√3 + 3) sin x + 2√3 cos x lies between – (2√3 + √15) and (2√3 + √15).
Answer» Suppose f(x) = (2√3 + 3)sin x + 2√3 cos x Here, A = 2√3, B = 2√3 + 3 and C = 0 – √[(2√3)² + (2√3 + 3)²] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[(2√3)² + (2√3 + 3)²] – √[12 + 12 + 9 + 12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[12 + 12 + 9 + 12√3] – √[33 + 12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[33 + 12√3] – √[15 + 12 + 6 + 12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[15 + 12 + 6 + 12√3] As we know that (12√3 + 6 < 12√5) because the value of √5 – √3 is more than 0.5 Therefore, if we replace, (12√3 + 6 with 12√5) the above inequality still holds. Therefore by rearranging the above expression √(15+12+12√5)we get, 2√3 + √15 – 2√3 + √15 ≤ (2√3 + 3) sin x + 2√3 cos x ≤ 2√3 + √15 Thus proved.

Prove that (2√3 + 3) sin x + 2√3 cos x lies between – (2√3 + √15) and (2√3 + √15).

Answer»

Suppose f(x) = (2√3 + 3)sin x + 2√3 cos x

Here, A = 2√3, B = 2√3 + 3 and C = 0

– √[(2√3)² + (2√3 + 3)²] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[(2√3)² + (2√3 + 3)²]
– √[12 + 12 + 9 + 12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[12 + 12 + 9 + 12√3]

– √[33 + 12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[33 + 12√3]

– √[15 + 12 + 6 + 12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[15 + 12 + 6 + 12√3]

As we know that (12√3 + 6 < 12√5) because the value of √5 – √3 is more than 0.5

Therefore, if we replace, (12√3 + 6 with 12√5) the above inequality still holds.

Therefore by rearranging the above expression √(15+12+12√5)we get, 2√3 + √15

– 2√3 + √15 ≤ (2√3 + 3) sin x + 2√3 cos x ≤ 2√3 + √15

Thus proved.

Discussion

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