If L = sin 2 π 16 - sin 2 π 8 and M = cos 2 π 16 - sin 2 π 8

L = sin π 16 + π 8 sin π 16 - π 8 = sin 3 π 16 · sin - π 16 = 1 2 cos 3 π 16 + π 16 - cos 3 π 16 - π 16 = 1 2 1 2 - cos π 8 M = cos π 16 + π 8 cos π 16 - π 8 = cos 3 π 16 · cos - π 16 = 1 2 cos 3 π 16 + π 16 + cos 3 π 16 - π 16 = 1 2 1 2 + cos π 8

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Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

MathematicsTrigonometric Ratios & IdentitiesAddition, Multiplication, Subtraction formula based,Multiple/Submultiple anglesEasy2 minPYQ_2020

MathematicsEasysingle choice

If $L = \sin^{2} (\frac{π}{16}) - \sin^{2} (\frac{π}{8})$ and $M = \cos^{2} (\frac{π}{16}) - \sin^{2} (\frac{π}{8})$

Options:

Answer:

Solution:

$L = \sin (\frac{π}{16} + \frac{π}{8}) \sin (\frac{π}{16} - \frac{π}{8})$
$= \sin \frac{3 π}{16} \cdot \sin (- \frac{π}{16})$

$= \frac{1}{2} (\cos (\frac{3 π}{16} + \frac{π}{16}) - \cos (\frac{3 π}{16} - \frac{π}{16})) = \frac{1}{2} (\frac{1}{\sqrt{2}} - \cos \frac{π}{8})$
$M = \cos (\frac{π}{16} + \frac{π}{8}) \cos (\frac{π}{16} - \frac{π}{8})$

$= \cos \frac{3 π}{16} \cdot \cos (- \frac{π}{16})$
$= \frac{1}{2} (\cos (\frac{3 π}{16} + \frac{π}{16}) + \cos (\frac{3 π}{16} - \frac{π}{16})) = \frac{1}{2} (\frac{1}{\sqrt{2}} + \cos \frac{π}{8})$

Stream:JEESubject:MathematicsTopic:Trigonometric Ratios & IdentitiesSubtopic:Addition, Multiplication, Subtraction formula based,Multiple/Submultiple angles

⏱ 2mℹ️ Source: PYQ_2020

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