Question

cos3A ×cos2A +sin4A×sin A= cos A ×cos 2A prove it

Answer 1

Answer:

$cos3A.cos2A+sin4A.sinA=cosA.cos2A$

Step-by-step explanation:

Formula used:

$cos(A+B)+cos(A-B)=2\:cosA\:cosB$

$cos(A-B)+cos(A+B)=2\:sinA\:sinB$

$cosC+cosD=2\:cos(\frac{C+D}{2})\:cos(\frac{C-D}{2})$

Now,

$cos3A.cos2A+sin4A.sinA$

$=\frac{1}{2}[2\:cos3A.cos2A+2\:sin4A.sinA]$

$=\frac{1}{2}[cos(3A+2A)+cos(3A-2A)+cos(4A-A)-cos(4A+A)]$

$=\frac{1}{2}[cos5A+cosA+cos3A-cos5A]$

$=\frac{1}{2}[cosA+cos3A]$

$=\frac{1}{2}[2\:cos(\frac{A+3A}{2}).cos(\frac{A-3A}{2})]$

$=cos2A.cos(-A)$

$=cos2A.cosA$

Answer 2

May its helps you....

3 formulas has been used in this question..

Firstly take LHS..

then put the formulas and you find...

LHS =RHS

Thank you