Question

Show that
Show that 2 sin2β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α.

Answer 1

$\large \mathfrak {\checkmark \: \pink{Question:-}}$

Show that :-

$\bf {2sin^2 \beta +}4cos( \alpha + \beta ) \: sin \alpha \: sin \beta +cos2( \alpha + \beta ) = cos2 \alpha$

$\large \mathfrak {\checkmark \: \pink{Solution:-}}$

$\bf LHS = 2sin^2 \beta + 4cos( \alpha + \beta ) \: sin \alpha \: sin \beta + cos2( \alpha + \beta )$

$\bf 2 {sin}^{2} \beta + 4(cos \alpha \: cos \beta - sin \alpha \: sin \beta )sin \alpha \: sin \beta + (cos2 \alpha cos2 \beta - sin2 \alpha sin2 \beta )$

$\bf 2 {sin}^{2} \beta + 4sin \alpha \: cos \alpha \: sin \beta \: cos \beta - 4 {sin}^{2} \alpha \: {sin}^{2} \beta + cos2 \alpha \: cos2 \beta - sin2 \alpha \: sin2 \beta$

$\bf 2 {sin}^{2} \beta + sin2 \alpha \: sin2 \beta - 4 {sin}^{2} \alpha \: {sin}^{2} \beta + cos2 \alpha \: cos2 \beta - sin2 \alpha \: sin2 \beta$

$\bf (1 - cos2 \beta ) - (2 {sin}^{2} \alpha )(2 {sin}^{2} \beta ) + cos2 \alpha \: cos2 \beta$

$\bf (1 - cos2 \beta ) - (1 - cos2 \alpha )(1 - cos2 \beta ) + cos2 \alpha \: cos2 \beta$

$\bf 1 - cos2 \beta + cos2 \alpha \: cos2 \beta - (1 - cos2 \beta - cos2 \alpha + cos2 \alpha \: cos2 \beta )$

$\bf 1-cos2 \beta +cos2 \alpha \: cos2 \beta -1+cos2 \beta + cos2 \alpha - cos2 \alpha \: cos2 \beta$

$\bf \cancel{1}-\cancel{cos2 \beta }+\cancel{cos2 \alpha ~cos2 \beta }-\cancel1+\cancel{cos2 \beta }+cos \alpha -\cancel{cos2 \alpha ~cos2 \beta }$

$\bf cos2 \alpha =RHS$

$\underline {\boxed {\bf {Hence, proved~!!}}}$

$\large \mathfrak {\checkmark \: \pink{Know \: more:-}}$

$\bf tan2A= \cfrac{2tanA}{1 - {tan}^{2} A}$

$\bf sinA=2sin \frac{A}{2} cos \frac{A}{2} = \cfrac{2tan \frac{A}{2} }{1 + tan^2 \frac{A}{2} }$

$\bf cosA= \cfrac{1 - {tan}^{2} \frac{A}{2} }{1 + {tan^2 \frac{A}{2} } }$

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Answer 2

LHS = 2 sin2β + 4 cos (α + β) sin α sin β + cos 2(α + β)

= 2 sin2β + 4 (cos α cos β – sin α sin β) sin α sin β + (cos 2α cos 2β – sin 2α sin 2β)

= 2 sin2β + 4 sin α cos α sin β cos β – 4 sin2α sin2β + cos 2α cos 2β – sin 2α sin 2β

= 2 sin2β + sin 2α sin 2β – 4 sin2α sin2β + cos 2α cos 2β – sin 2α sin 2β

= (1 – cos 2β) – (2 sin2α) (2 sin2β) + cos 2α cos 2β

= (1 – cos 2β) – (1 – cos 2α) (1 – cos 2β) + cos 2α cos 2β

= cos 2α

= RHS

Therefore, 2 sin2β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α

$LHS = 2 sin2β + 4 cos (α + β) sin α sin β + cos 2(α + β)= 2 sin2β + 4 (cos α cos β – sin α sin β) sin α sin β + (cos 2α cos 2β – sin 2α sin 2β)= 2 sin2β + 4 sin α cos α sin β cos β – 4 sin2α sin2β + cos 2α cos 2β – sin 2α sin 2β= 2 sin2β + sin 2α sin 2β – 4 sin2α sin2β + cos 2α cos 2β – sin 2α sin 2β= (1 – cos 2β) – (2 sin2α) (2 sin2β) + cos 2α cos 2β= (1 – cos 2β) – (1 – cos 2α) (1 – cos 2β) + cos 2α cos 2β= cos 2α= RHSTherefore, 2 sin2β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α$