Question

The maximum value of (cos α₁)(cos α₂)...(cosαₙ) under the restrictions 0 ≤ α₁,α₂,...,αₙ ≤ π/2 and (cot α₁)(cot α₂)...(cot αₙ) = 1 is

(a) $\dfrac{1}{2^{\frac{n}{2} } }$


(b) $\dfrac{1}{2^{n} }$


(c) $\dfrac{1}{2n}$


(d) 1

Answer 1

Given :  0 ≤ α₁,α₂,...,αₙ ≤ π/2 and (cot α₁)(cot α₂)...(cot αₙ) = 1  

To find : The maximum value of (cos α₁)(cos α₂)...(cosαₙ)

Solution:

(cot α₁)(cot α₂)...(cot αₙ) = 1

=> (cos α₁)(cos α₂)...(cosαₙ)  /  (Sin α₁)(Sin α₂)...(Sinαₙ)  = 1

=> (cos α₁)(cos α₂)...(cosαₙ) = (Sin α₁)(Sin α₂)...(Sinαₙ)

multiplying both sides by  (cos α₁)(cos α₂)...(cosαₙ)

=> ( (cos α₁)(cos α₂)...(cosαₙ) )² =  (cos α₁)(cos α₂)...(cosαₙ)  (Sin α₁)(Sin α₂)...(Sinαₙ)

=> ( (cos α₁)(cos α₂)...(cosαₙ) )²  = (cos α₁Sin α₁)(cos α₂Sin α₂) ......(cosαₙSinαₙ)

using Sin2x = 2SinxCosx

=>  ( (cos α₁)(cos α₂)...(cosαₙ) )²  = (Sin 2α₁ / 2)(Sin 2α₂ / 2) ......(Sin2αₙ / 2)

=>  ( (cos α₁)(cos α₂)...(cosαₙ) )²  = (1/2ⁿ) (Sin 2α₁ )(Sin 2α₂) ......(Sin2αₙ)

as we know

0 ≤ α₁,α₂,...,αₙ ≤ π/2

=>  0 ≤ 2α₁,2α₂,...,2αₙ ≤ π

Sin is + ve from

Maximum value of (Sin 2α₁ )(Sin 2α₂) ......(Sin2αₙ) = 1

=>  ( (cos α₁)(cos α₂)...(cosαₙ) )²   ≤ (1/2ⁿ)

=>  (cos α₁)(cos α₂)...(cosαₙ)  ≤  $\frac{1}{2^{ \frac{n}{2} }}$

Hence The maximum value of (cos α₁)(cos α₂)...(cosαₙ) under the restrictions 0 ≤ α₁,α₂,...,αₙ ≤ π/2 and (cot α₁)(cot α₂)...(cot αₙ) = 1 is     ≤  $\frac{1}{2^{ \frac{n}{2} }}$

option a is correct

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Answer 2

Answer:

Option a is the answer of this question ok