Verified Solution • 2025–26 Session moderate

QJEEM20Ma1475: JEE Main 2020 Mathematics - Class 11 JEEM20Ma1475 Solution: If L = sin 2 - sin 2 and M = cos 2 - sin 2 , then :

If L = sin2 (π16)\left( {{\pi \over {16}}} \right) - sin2 (π8)\left( {{\pi \over {8}}} \right) and
M = cos2 (π16)\left( {{\pi \over {16}}} \right) - sin2 (π8)\left( {{\pi \over {8}}} \right) , then :

Multiple Choice Options

  • A L = −122+12cos⁡π8- {1 \over {2\sqrt 2 }} + {1 \over 2}\cos {\pi \over 8}
  • B M = 122+12cos⁡π8{1 \over {2\sqrt 2 }} + {1 \over 2}\cos {\pi \over 8}
  • C M = 142+14cos⁡π8{1 \over {4\sqrt 2 }} + {1 \over 4}\cos {\pi \over 8}
  • D L = 142−14cos⁡π8{1 \over {4\sqrt 2 }} - {1 \over 4}\cos {\pi \over 8}

Step-by-Step Verified Solution

We will use here those two formulas,

sin2 θ\theta = 1−cos⁡2θ2{{1 - \cos 2\theta } \over 2} and cos2 θ\theta = 1+cos⁡2θ2{{1 + \cos 2\theta } \over 2}

L = sin2 (π16)\left( {{\pi \over {16}}} \right) - sin2 (π8)\left( {{\pi \over {8}}} \right)

⇒\Rightarrow L = (1−cos⁡(π8)2)\left( {{{1 - \cos \left( {{\pi \over 8}} \right)} \over 2}} \right) - (1−cos⁡(π4)2)\left( {{{1 - \cos \left( {{\pi \over 4}} \right)} \over 2}} \right)

⇒\Rightarrow L = 12(cos⁡(π4)−cos⁡(π8)){1 \over 2}\left( {\cos \left( {{\pi \over 4}} \right) - \cos \left( {{\pi \over 8}} \right)} \right)

⇒\Rightarrow L = 122−12cos⁡(π8){1 \over {2\sqrt 2 }} - {1 \over 2}\cos \left( {{\pi \over 8}} \right)

M = cos2 (π16)\left( {{\pi \over {16}}} \right) - sin2 (π8)\left( {{\pi \over {8}}} \right)

⇒\Rightarrow M = (1+cos⁡(π8)2)\left( {{{1 + \cos \left( {{\pi \over 8}} \right)} \over 2}} \right) - (1−cos⁡(π4)2)\left( {{{1 - \cos \left( {{\pi \over 4}} \right)} \over 2}} \right)

⇒\Rightarrow M = 122+12cos⁡π8{1 \over {2\sqrt 2 }} + {1 \over 2}\cos {\pi \over 8}

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