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JEE Main 2024
Trigonometry
Trigonometric Ratio and Identites
Medium

Question

If L = sin 2 (π16)\left( {{\pi \over {16}}} \right) - sin 2 (π8)\left( {{\pi \over {8}}} \right) and M = cos 2 (π16)\left( {{\pi \over {16}}} \right) - sin 2 (π8)\left( {{\pi \over {8}}} \right), then :

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Solution

We will use here those two formulas, sin 2 θ\theta = 1cos2θ2{{1 - \cos 2\theta } \over 2} and cos 2 θ\theta = 1+cos2θ2{{1 + \cos 2\theta } \over 2} L = sin 2 (π16)\left( {{\pi \over {16}}} \right) - sin 2 (π8)\left( {{\pi \over {8}}} \right) \Rightarrow L = (1cos(π8)2)\left( {{{1 - \cos \left( {{\pi \over 8}} \right)} \over 2}} \right) - (1cos(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 = 12212cos(π8){1 \over {2\sqrt 2 }} - {1 \over 2}\cos \left( {{\pi \over 8}} \right) M = cos 2 (π16)\left( {{\pi \over {16}}} \right) - sin 2 (π8)\left( {{\pi \over {8}}} \right) \Rightarrow M = (1+cos(π8)2)\left( {{{1 + \cos \left( {{\pi \over 8}} \right)} \over 2}} \right) - (1cos(π4)2)\left( {{{1 - \cos \left( {{\pi \over 4}} \right)} \over 2}} \right) \Rightarrow M = 122+12cosπ8{1 \over {2\sqrt 2 }} + {1 \over 2}\cos {\pi \over 8}

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