Verified Solution • 2025–26 Session moderate

QJEEM19Maaae6: JEE Main 2019 Mathematics - Class 11 JEEM19Maaae6 Solution: If sin 4 + 4 cos 4 + 2 = 4 sin cos ; , [0, ], then cos( + )…

If sin4 α\alpha + 4 cos4 β\beta + 2 = 4 2\sqrt 2 sin α\alpha cos β\beta ; α\alpha , β\beta ∈\in [0, π\pi ],
then cos( α\alpha + β\beta ) −- cos( α\alpha −- β\beta ) is equal to :

Multiple Choice Options

  • A −2- \sqrt 2
  • B 0
  • C −- 1
  • D 2\sqrt 2

Step-by-Step Verified Solution

A.M. ≥\ge G.M.

sin⁡4α+4cos⁡4β+1+14≥(sin⁡4α.4cos⁡4β.1.1)14{{{{\sin }^4}\alpha + 4{{\cos }^4}\beta + 1 + 1} \over 4} \ge {\left( {{{\sin }^4}\alpha .4{{\cos }^4}\beta .1.1} \right)^{{1 \over 4}}}

sin4 α\alpha + 4 cos2 β\beta + 2 ≥\ge 4 2\sqrt 2 sin α\alpha cos β\beta

Given that sin4 α\alpha + 4cos4 β\beta + 2 = 4 2\sqrt 2 sin α\alpha cos β\beta

⇒\Rightarrow   A.M.=G.M. ⇒\Rightarrow sin4 α\alpha = 1 = 4 cos4 β\beta

sin α\alpha = 1, cos β\beta = ±\pm 12{1 \over {\sqrt 2 }}

⇒\Rightarrow    sin β\beta = 12{1 \over {\sqrt 2 }} as β\beta ∈\in [0, π\pi ]

cos( α\alpha + β\beta ) −- cos ( α\alpha −- β\beta ) = −- 2 sin α\alpha β\beta

= −2- \sqrt 2

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