Question

please I need help its urgent ​

Answer 1

Step-by-step explanation:

$we \: know \: that \\ cos2 \alpha = 1 - 2 {sin}^{2} \alpha \\ 2 {sin}^{2} \alpha = 1 - cos2 \alpha \\ now$

${sin}^{2} ( \dfrac{\pi}{8} + \dfrac{x}{2} ) = \dfrac{1}{2} \: 2 {sin}^{2} ( \dfrac{\pi}{8} + \dfrac{x}{2} ) \\ = \dfrac{1}{2} ( \: 1 - cos2( \dfrac{\pi}{8} + \dfrac{x}{2} ) \: ) \\ = \dfrac{1}{2} (1 - cos( \dfrac{\pi}{4} + \dfrac{x}{2} ) \: )$

$similarly \\ {sin}^{2} (\dfrac{\pi}{8} - \dfrac{x}{2} ) = \dfrac{1}{2} ( \: 1 - cos( \dfrac{\pi}{4} - x) \: )$

$lhs = {sin}^{2}(\dfrac{\pi}{8} +\dfrac{x}{2} )- {sin}^{2} (\dfrac{\pi}{8}-\dfrac{x}{2})$

=1+cos(π/4 +x) - ( 1 - cos(π/4-x) )

=1+cos (π/4+x)-1+cos(π/4-x)

=cos(π/4+x)+cos(π/4-x)

=cosπ/4 cosx- sinπ/4 sinx +cosπ/4 cosx+ sinπ/4 sinx

= 2 cosπ/4 cosx

= 2 (1/√2) cosx

=√2 cosx

Answer 2

HOPE THIS HELPS YOU....