Pls answer this...
If u answer without explanation or spam,
then u r a DONKEY...
Answers
RHS = cos3A/(2cos2A - 1)
we know, cos3x = 4cos³x - 3cosx
cos2x = 2cos²x - 1
so, cos3A/(2cos2A - 1) = (4cos³A - 3cosA)/{2(2cos²A - 1) - 1}
= (4cos³A - 3cosA)/(4cos²A - 3 )
= cosA(4cos²A - 3)/(4cos²A - 3)
= cosA = LHS [ hence proved]
hopes it's helps uh !!!!!!!!!!!!!!!!!
❤️
Step-by-step explanation:
cosA = cos3A/(2cos2A-1)
this question is to show that hence,it can be proved in two ways
(1) solve for LHS and show its equal to RHS
(2)solve for RHS and show its equal to LHS
(1) cos A
multiply and divide by (2cos2A-1)
= {cosA (2cos2A-1) }/(2cos2A-1)
={2cosA.cos2A-cosA}/(2cos2A-1)
apply cosx.cosy formula
= {(1/2){2(cosA-2A)+2(cosA+2A)}-cosA}/(2cos2A-1)
you know cos (-x)= cosx
={cos A+cos3A-cosA}/(2cos2A-1)
=cos3A/(2cos2A-1)
hence proved LHS=RHS
(2) this method you can find in n no.of books
any queries?
thank you have a nice day