Question

show that (sintheta+costheta)^2-(sintheta-costheta)^2=2sin2theta​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO PROVE

$\sf{ {( \sin \theta + \cos \theta)}^{2} - \: {( \sin \theta - \cos \theta)}^{2} = 2 \sin 2\theta}$

FORMULA TO BE IMPLEMENTED

$1. \: \: \sf{ {a}^{2} - {b}^{2} = (a + b)(a - b) \: }$

$2. \: \: \sf { \: \sf{ {2\sin \theta \cos \theta} = \sin 2\theta} }$

CALCULATION

LHS

$= \sf{ {( \sin \theta + \cos \theta)}^{2} - \: {( \sin \theta + \cos \theta)}^{2} }$

$= \sf{ \bigg[ {( \sin \theta + \cos \theta)} + \: {( \sin \theta - \cos \theta)}\bigg] \times \bigg[ {( \sin \theta + \cos \theta)} - \: {( \sin \theta - \cos \theta)}\bigg] \: }$

$= \sf{ \bigg[ { \sin \theta + \cos \theta} + \: { \sin \theta - \cos \theta}\bigg] \times \bigg[ { \sin \theta + \cos \theta} - \: {\sin \theta + \cos \theta}\bigg] \: }$

$= \sf { \sf{ {2\sin \theta \times 2 \cos \theta} }}$

$= \sf { \sf{ {4\sin \theta \cos \theta} }}$

$= \sf { \sf{ {2 \times 2\sin \theta \cos \theta} }}$

$= \sf{ {2 \sin 2\theta }}$

= RHS

Hence proved

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