Question

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

Answer 1

━━━━━━━━━━━━━━━━━━━━━━━━━

$\Large\fbox{\color{purple}{QUESTION}}$

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

━━━━━━━━━━━━━━━━━━━━━━━━━

$\bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }$

➡️LHS = 2 sin^2β + 4 cos (α + β) sin α sin β + cos 2(α + β)

➡️= 2 sin^2β + 4 (cos α cos β – sin α sin β) sin α sin β + (cos 2α cos 2β – sin 2α sin 2β)

➡️= 2 sin^2β + 4 sin α cos α sin β cos β – 4 sin^2α sin^2β + cos 2α cos 2β – sin 2α sin 2β

➡️= 2 sin^2β + sin 2α sin 2β – 4 sin^2α sin^2β + cos 2α cos 2β – sin 2α sin 2β

➡️= (1 – cos 2β) – (2 sin^2α) (2 sin^2β) + cos 2α cos 2β

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

➡️= cos 2α

➡️= RHS

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

$\Large\mathcal\green{FOLLOW \: ME}$

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

SOLUTION:-

➡️LHS = 2 sin^2β + 4 cos (α + β) sin α sin β + cos 2(α + β)

➡️= 2 sin^2β + 4 (cos α cos β – sin α sin β) sin α sin β + (cos 2α cos 2β – sin 2α sin 2β)

➡️= 2 sin^2β + 4 sin α cos α sin β cos β – 4 sin^2α sin^2β + cos 2α cos 2β – sin 2α sin 2β

➡️= 2 sin^2β + sin 2α sin 2β – 4 sin^2α sin^2β + cos 2α cos 2β – sin 2α sin 2β

➡️= (1 – cos 2β) – (2 sin^2α) (2 sin^2β) + cos 2α cos 2β

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

➡️= cos 2α

➡️= RHS

➡️Therefore, 2 sin^2β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α✔

Step-by-step explanation:

HOPE IT HELP YOU ✌✌