Question

Prove the following !!!!

( 1 - 2sin²A cos²A ) (sin²A - cos²A) = sin⁸A - cos⁸A


help needed !!!!

Answer 1

Hey there !

starting with RHS : 
=> sin⁸A - cos⁸A

we know that :-
sin⁸A = (sin⁴A)² 
cos⁸A = (cos⁴A)²

This can be written as :-

 => (sin⁴A)² - (cos⁴A)²

Now this is in the form of an identity : a² - b² = (a+b) ( a - b)

 => (sin⁴A + cos⁴A) ( sin⁴A - cos⁴A)

sin⁴A  = (sin²A)²
cos⁴A = (cos²A)²

=>  (sin²A)² + (cos²A)² (( sin⁴A - cos⁴A))


 (sin²A)² + (cos²A)² can be in the identity : a² + b² = (a+b)² - 2ab

  [    (sin²A)² + (cos²A)² =  (sin²A + cos²A)² - 2sin²A cos²A ]



 => [(sin²A + cos²A)² - 2sin²A cos²A ] (( sin⁴A - cos⁴A))


Now ,
sin⁴A - cos⁴A 
this can be written in the form of the identity a² - b² = (a+b) (a -b)

sin⁴A  = (sin²A)²
cos⁴A = (cos²A)²

 sin⁴A - cos⁴A = (sin²A + cos²A) (sin²A - cos²A)

=> => [(sin²A + cos²A)² - 2sin²A cos²A ] (sin²A + cos²A) (sin²A - cos²A)

we know that ,
sin²A + cos²A = 1              [ by identity ]

hence,

=> [ (1)² - 2sin²A cos²A ] (1) ×(sin²A - cos²A)

=> ( 1 - 2sin²A cos²A ) (sin²A - cos²A)

=>LHS

Answer 2

Given, RHS = sin^8A - cos^8A  can be written as

= (sin^4A)^2 - (cos^4A)^2 

We know that  a^2 - b^2 = (a+b) * (a-b). 

= (sin^4A + cos^4A) * (sin^4A - cos^4A) 

= (sin^2A)^2 + (cos^2A)^2) * ((sin^2A)^2 - (cos^2A)^2)

= (sin^2A+cos^2A)^2 - 2sin^2Acos^2A) * (sin^2A+cos^2A) * (sin^2A-cos^2A)

We know that (sin^2A + cos^2A) = 1 

= ((1)^2 - 2sin^2Acos^2A)) * 1 * (sin^2A - cos^2A) 

= (1 - 2sin^2Acos^2A) * (sin^2A - cos^2A) 

= (sin^2A - cos^2A) * (1 - 2sin^2Acos^2A).

LHS = RHS.

Hope this helps!