If L=sin^2(π)/16-sin^2(π)/8 and M=cos^2(π)/16-sin^2(π)/8

Correct answer: (D) M=1/(2√2)+1/2cos(π)/8. L=sin(π)/16+(π)/8sin(π)/16-(π)/8 =sin(3π)/16·sin-(π)/16 =1/2cos(3π)/16+(π)/16-cos(3π)/16-(π)/16=1/2(1√2-cosπ8M=cosπ16+π8cosπ16-π8=cos3π16·cos-π16=1)/2cos(3π)/16+(π)/16+cos(3π)/16-(π)/16=1/21√2+cos(π)/8

Q32 / 75 · JEE Main 2020 (05 Sep Shift 2) · Mathematics

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

A
$L = - \frac{1}{2 \sqrt{2}} + \frac{1}{2} \cos \frac{π}{8}$
B
$L = \frac{1}{4 \sqrt{2}} - \frac{1}{4} \cos \frac{π}{8}$
C
$M = \frac{1}{4 \sqrt{2}} + \frac{1}{4} \cos \frac{π}{8}$
Correct answer
D
$M = \frac{1}{2 \sqrt{2}} + \frac{1}{2} \cos \frac{π}{8}$

Solution

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

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

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

Hence, the correct answer is option: 'D'.

If L=sin2π16-sin2π8 and M=cos2π16-sin2π8

Solution

Relevant Questions

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