Question

Prove that: 2(sin⁶θ+cos⁶θ) - 3(sin⁴θ+cos⁴θ)+1 = 0

Answer 1

GIVEN:
2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1 = 0

LHS = 2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1

=2[(sin²θ)³+(cos²θ)³] - 3sin⁴θ-3cos⁴θ+1
=2(sin²θ+cos²θ)[(sin²θ)²+(cos²θ)²-sin²θcos²θ)] - 3sin⁴θ -3cos⁴θ+1

[a³ + b³ = (a+b) (a²+b² -a-b]

=2(1)[sin⁴θ+cos⁴θ - sin²θ cos²θ] - 3sin⁴θ - 3cos⁴θ +1                              
 [sin²θ+cos²θ=1]

=2sin⁴θ+2cos⁴θ-2sin²θcos²θ -3sin⁴θ - 3cos⁴θ +1

= 2sin⁴θ -3sin⁴θ +2cos⁴θ - 3cos⁴θ - 2sin²θcos²θ +1
= - sin⁴θ - cos⁴θ - 2sin² θcos²θ + 1
=1 - [sin⁴θ+cos⁴θ+2sin²θ(cos²θ)]
=1 - [(sin²θ)²+(cos²θ)²+2sin²θcos²θ]

=1 - (sin²θ+cos²θ)²

[ a² + b² +2ab = (a+b)² ]

=1 - (1)²
LHS = 1-1 =0

LHS = RHS
Hence, proved  2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1=0

HOPE THIS WILL HELP YOU..

Answer 2

Trigonometry,

Let A instead of theta
2(sin^6A+cos^6A)-3(sin^4A+cos^4A)

Have to prove this = 1

Now
LHS = 2(sin^6A+cos^6A)-3(sin^4A+cos^4A)+1

=2{(sin^2A)^3+(cos^2A)^3}-3{(sin²A+cos²A)-2sin²A.cos²A}+1

= 2{(sin²A+cos²A)(sin⁴A-sin²A.cos²A+cos⁴A)}-3(1-2sin²A.cos²A)+1

= 2{(sin²A + cos²A)-2sin²A.cos²A -sin²A.cos²A}-3+6sin²A.cos²A+1

= 2(1-3sin²A.cos²A)-3+6sin²A.cos²A+1

= 2-6sin²A.cos²A - 3+6sin²A.cos2A+1

= -1+1

= 0 = RHS [proved]

That's it
Hope it helped (･ิω･ิ)