Question

prove that: sin6A+cos6A =1-3sin2A.cos2A

Answer 1

Hey ;!!

Cos^6A + sin^6A

=(Cos³A)² +( Sin³A)²

= (cos²A + sin²A ) ( Cos⁴A+ Sin⁴A - Cos²A×Sin²A)


= [°•° a³ + b³ = (a+b ) ( a²+ b²-ab)]

= (1 )(cos²A )² + (sin²A)² - 2cos²A× sin²A - cos²A× sin²A

= [ a²+ b² =[ (a+b)² -2ab ]

= 1²- 3sin²A×cos²A

= 1 - 3sin²A×cos²A RHS prooved

hope it helps !!!

#Rajukumar111@@@@

Answer 2

Hello friend,

Here's your answer:

As Cos^6A + sin^6A
=>(Cos³A)² +( Sin³A)²
=> (cos²A + sin²A ) ( Cos⁴A+ Sin⁴A - Cos²A×Sin²A)
=> Therefore [ a³ + b³ = (a+b ) ( a²+ b²-ab)]
=> (1 )(cos²A )² + (sin²A)² - 2cos²A× sin²A - cos²A× sin²A 
=> [ a²+ b² =[ (a+b)² -2ab ]
=> 1²- 3sin²A×cos²A
=> 1 - 3sin²A×cos²A

Hope my answer helps you.

Harith 
Maths Aryabhatta