If L-sin2 -16 -sin2 -8 and M-cos2 -16 -sin2 -8 - then-

L= sin^2 ( π16 ) sin^2 ( π8 ) ( ∵sin^2 θ = 1 cos2θ2) L= ( 1 cosπ82) ( 1 cosπ42) nbsp; L= 12 [ cosπ8 cosπ4 ] L=12√(2) 12cosπ8 M=cos^2 ( π16 ) sin^2 ...

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Byju's Answer

Standard XII

Mathematics

Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

If L=sin2 π16...

Question

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

M = \frac{1}{2 \sqrt{2}} + \frac{1}{2} cos \frac{π}{8}

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M = \frac{1}{4 \sqrt{2}} + \frac{1}{4} cos \frac{π}{8}

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L = - \frac{1}{2 \sqrt{2}} + \frac{1}{2} cos \frac{π}{8}

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L = \frac{1}{4 \sqrt{2}} - \frac{1}{4} cos \frac{π}{8}

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Solution

The correct option is A $M = \frac{1}{2 \sqrt{2}} + \frac{1}{2} cos \frac{π}{8}$
$L = {sin}^{2} (\frac{π}{16}) - {sin}^{2} (\frac{π}{8})$
$(∵ {sin}^{2} θ = \frac{1 - cos 2 θ}{2})$
$L = ⎛ ⎜ ⎜ ⎝ \frac{1 - cos \frac{π}{8}}{2} ⎞ ⎟ ⎟ ⎠ - ⎛ ⎜ ⎜ ⎝ \frac{1 - cos \frac{π}{4}}{2} ⎞ ⎟ ⎟ ⎠$
$L = \frac{- 1}{2} [cos \frac{π}{8} - cos \frac{π}{4}]$
$L = \frac{1}{2 \sqrt{2}} - \frac{1}{2} cos \frac{π}{8}$

$M = {cos}^{2} (\frac{π}{16}) - {sin}^{2} (\frac{π}{8})$
$M = ⎛ ⎜ ⎜ ⎝ \frac{1 + cos \frac{π}{8}}{2} ⎞ ⎟ ⎟ ⎠ - ⎛ ⎜ ⎜ ⎝ \frac{1 - cos \frac{π}{4}}{2} ⎞ ⎟ ⎟ ⎠$
$M = \frac{1}{2} [cos \frac{π}{4} + cos \frac{π}{8}]$
$M = \frac{1}{2 \sqrt{2}} + \frac{1}{2} cos \frac{π}{8}$

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Similar questions

Q. If

\frac{sin θ}{{sin}^{2} (\frac{π}{8} + \frac{θ}{2}) - {sin}^{2} (\frac{π}{8} - \frac{θ}{2})} = k (θ \neq 2 n π)

, then value of

2 k^{2}

Q. If

{⎡ ⎣ \begin{matrix} cos \frac{2 π}{3} & - sin \frac{2 π}{3} sin \frac{2 π}{3} & cos \frac{2 π}{3} \end{matrix} ⎤ ⎦}^{k} = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

, then least value of

k

equals

(k \neq 0)

Q. Solve the following equations.

s i n^{2} (\frac{π}{8} + x) = s i n x + s i n^{2} (\frac{π}{8} - x) .

Find the value of $s i n^{2} \frac{π}{8}$ + $s i n^{2} \frac{π}{4}$

Q. lf

f (x) = {sin}^{2} (\frac{π}{8} + \frac{x}{2}) - {sin}^{2} (\frac{π}{8} - \frac{x}{2})

, then the period of

f

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If L=sin2(π16)−sin2(π8) and M=cos2(π16)−sin2(π8), then:

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