2(sin^6x + cos^6x) - 3(sin^4x + cos^4x) + 1

Question

Solve:  2(sin⁶x + cos⁶x) - 3(sin⁴x + cos⁴x) + 1

Answer 1

LHS = 2(sin⁶x + cos⁶x) − 3(sin⁴x + cos⁴x) + 1

= 2[(sin²x)³ + (cos²x)³] − 3(sin⁴x + cos⁴x) + 1

= 2[(sin²x + cos²x)(sin⁴x − sin²x cos²x + cos⁴x)] − 3(sin⁴x + cos⁴x) + 1

[∵ a³ + b³ = (a + b)(a² − ab + b²)]

= −[sin⁴x + cos⁴x + 2sin²xcos²x] + 1

= −[sin²x + cos²x] + 1

= -1 + 1

= 0

= RHS

Hence Proved.

Answer 2

---------------------------

Answer 3

We know that

=> sin²A+cos²A=1
=> 1-sin²A=cos²A
=> 1-cos²A=sin²A
Applying these and solving we have:
=2(sin⁶A+cos⁶A)-3(sin⁴A+cos⁴A)+1

=2sin⁶A+2cos⁶A-3sin⁴A-3cos⁴A+sin²A+cos²A

=2sin⁶A+2cos⁶A-2sin⁴A-2cos⁴A-sin⁴A-cos⁴A+sin²A+cos²A

=2sin⁶A-2sin⁴A+2cos⁶A-2cos⁴A+sin²A-sin⁴A+cos²A-cos⁴A

=-2sin⁴A(1-sin²A)-2cos⁴A(1-cos²A)+sin²A(1-sin²A)+cos²A(1-cos²A)

=-2sin⁴Acos²A-2cos⁴Asin²A+sin²Acos²A+cos²Asin²A

=-2sin²Acos²A(sin²A+cos²A)+2sin²Acos²A

=-2sin²Acos²A+2sin²Acos²A

=0

Hence,
2(sin⁶A+cos⁶A)-3(sin⁴A+cos⁴A)+1=0

Comments

Nice answer