Question

prove 2(sin^6theta+cos^6theta)-3(sin^4theta+cos^4theta)+1=1

PLEASE SHOW HOLE process

Answer 1

Hey !!!

here is ur answer !!!

2 ( sin^6 ¢+ cos^6¢) - 3 ( sin⁴ ¢ + cos⁴ ¢ ) +1

=> 2{ ( sin² ¢) ³ + (cos²¢)³ } -3 { ( sin²¢)² +( cos²¢)² } +1

= > 2( sin²¢ + cos²¢ )(sin⁴¢ + cos⁴¢ -sin²¢ × cos²¢ } -3{ ( 3sin²¢)² +( cos²¢)² - 2sin²¢ × cos²¢ +1

=> 2{( sin²¢ )² +( cos²¢ )² - sin²¢ ×cos²¢ -3(1-2sin²¢×cos²¢)+1

=> 2{(sin²¢ +cos²¢ )² - 2 sin²¢ × cos²¢ -sin²¢ × cos²¢ } -3(1 - 2sun²¢ ×cos²¢ ) +1

=> 2(1 - 3sin²¢ cos²¢ )} -3(1 - 2sin²¢ ×cos²¢ )+1

=] 2( 1 - 6sin²¢ ×cos²¢ - 3 + 6sin²¢×cos²¢ +1

=> 2-6sin²¢ ×cos²¢ -3 + 6sin²¢ ×cos²¢ +1 .

=> 3 -3 =0 ...question is wrong dear..please check out. ..

hope it helps !!.

#Rajukumar111

Answer 2

2((sin²)³+(cos²)³)-3((sin²)²+(cos²)²)+1
as by identity
a³+b³=(a²+b²-ab)(a+b)
and as a²+ b²=(a+b)²-2ab
=2(sin²+cos²-sin²cos²)(sin²+cos²) -
3((sin²+cos²)²-2sin²cos²)+1
=2(1-sin²cos²) - 3(1-2sin²cos²)+1
=2-2sin²cos²-3-6sin²cos²+1
=-8sin²cos²-1+1
=4(-2sin²cos²)
as identity
(a+b)²=a²+b²+2ab. => -2ab=a²+b²-(a+b)²
so,
=4(sin⁴+cos⁴-(sin²+cos²))