Question

If \({\rm{\;si}}{{\rm{n}}^4}{\rm{\alpha }} + 4{\rm{co}}{{\rm{s}}^4}{\rm{\beta }} + 2 = 4\sqrt 2 {\rm{sin\alpha \;cos\beta }};{\rm{\;\alpha }},{\rm{\beta }} \in \left[ {0,{\rm{\pi }}} \right],{\rm{}}\) then cos (α + β) – cos (α – β) is equal to
1. \(- 1\)
2. \(\sqrt 2\)
3. \(- \sqrt 2\)
4. 0

Answer 1

Correct Answer - Option 3 : \(- \sqrt 2\)

By applying AM ≥ GM inequality, on the numbers sin⁴ α, 4 cos⁴ β, 1 and 1, we get

\(\frac{{{\rm{si}}{{\rm{n}}^4}{\rm{\alpha }} + 4{\rm{co}}{{\rm{s}}^4}{\rm{\beta }} + 2}}{4} \ge {\left( {\left( {{\rm{si}}{{\rm{n}}^4}{\rm{\alpha }}} \right)\left( {4{\rm{co}}{{\rm{s}}^4}{\rm{\beta }}} \right)\cdot1\cdot1} \right)^{\frac{1}{4}}}\)

\(= {\rm{si}}{{\rm{n}}^4}{\rm{\alpha }} + 4{\rm{co}}{{\rm{s}}^4}{\rm{\beta }} + 2 \ge 4\sqrt 2 {\rm{sin\alpha \;cos\beta }}\)

But, it is given that

\({\rm{si}}{{\rm{n}}^4}{\rm{\alpha }} + 4{\rm{co}}{{\rm{s}}^4}{\rm{\beta }} + 2 = 4\sqrt 2 {\rm{sin\alpha \;cos\beta }}\)

So sin⁴ α = 4 cos⁴ β = 1

[In AM ≥ GM, equality holds when an given positive quantities are equal.]

\(\Rightarrow {\rm{\;sin\alpha }} = 1{\rm{\;and\;sin\beta }} = \frac{1}{{\sqrt 2 }}{\rm{\;}}\)    ----(2)

[∵ α, β ∈ [0, π]]

Now, cos (α + β) – cos (α – β)

= -2 sin α sin β

\(\left[\because {{\rm{cosC}} - {\rm{cosD}} = 2{\rm{sin}}\frac{{{\rm{C}} + {\rm{D}}}}{2}{\rm{sin}}\frac{{{\rm{D}} - {\rm{C}}}}{2}} \right]\)

\(= - 2 \times 1 \times \frac{1}{{\sqrt 2 }}{\rm{\;}}\) [from Eq,(1)]

\(= - \sqrt 2\)