Question

$\footnotesize{\underline{\textsf{\textbf{\red{Q\;U\;E\;S\;T\;I\;O\;N\;\;\;:-}}}}}$
Prove the following
i) $\sf 2sin^2\dfrac{\pi}{6}+cosec^2\dfrac{7\pi}{6}cos^2\dfrac{\pi}{3}=\dfrac{3}{2}$
ii) $\sf cos^22x-cos^26x=sin4x\;sin8x$

Answer 1

Step-by-step explanation:

$\rm\:\bold{i.}$

We have,

$\rm 2sin^2\dfrac{\pi}{6}+cosec^2\dfrac{7\pi}{6}cos^2\dfrac{\pi}{3}$

$\rm = 2 \bigg(\frac{1}{2} \bigg)^{2} +cosec^2 \bigg(\pi + \frac{\pi}{6} \bigg). \bigg(\frac{1}{2} \bigg)^{2}$

$\rm = 2 \times \frac{1}{4} + \frac{1}{4} \times cosec^2 \bigg( \frac{\pi}{6} \bigg)$

$\rm = \frac{1}{2} + \frac{1}{4} \times (2)^2$

$\rm = \frac{1}{2} + 1$

$\rm = \frac{3}{2}$

$\rm\:\bold{ii.}$

We have,

$\rm cos^2(2x)-cos^2(6x)$

$\rm = \{cos(2x)-cos(6x) \} \{ cos(2x) + cos(6x) \}$

$\rm = \bigg\{ - 2 \sin \bigg( \frac{2x + 6x}{2} \bigg) \sin \bigg( \frac{2x - 6x}{2} \bigg) \bigg\} \bigg\{ 2cos \bigg( \frac{2x + 6x}{2} \bigg) cos \bigg( \frac{2x - 6x}{2} \bigg)\bigg \}$

$\rm = \bigg\{ - 2 \sin \bigg( \frac{ 8x}{2} \bigg) \sin \bigg( \frac{ - 4x}{2} \bigg) \bigg\} \bigg\{ 2cos \bigg( \frac{ 8x}{2} \bigg) cos \bigg( \frac{ - 4x}{2} \bigg)\bigg \}$

$\rm = \bigg\{ 2 \sin \bigg( \frac{ 8x}{2} \bigg) \sin \bigg( \frac{ 4x}{2} \bigg) \bigg\} \bigg\{ 2cos \bigg( \frac{ 8x}{2} \bigg) cos \bigg( \frac{ 4x}{2} \bigg)\bigg \}$

$\rm = 2 \sin ( 4x) \sin ( 2x ) . 2cos( 4x) cos ( 2x )$

$\rm = 2 \sin ( 4x) cos( 4x).2 \sin ( 2x ) cos ( 2x )$

$\rm = \sin ( 8x) \sin ( 4x )$

Answer 2

Answer:

Hey

Step-by-step explanation:

you are a good and quality answer person..u are intelligent

please see my first 4 mathematics Question

you will soon become a moderator