Question

Prove that
sin³a - cos³a = (sin²a - cos²a) (1-2sin² cos²a)

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Answer 1

Answer:

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

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