Question

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

Answer 1

Hey, please look at the attachment :) 

Answer 2

Solution:

Given

$2(sin^{6}A+cos^{6})-3(sin^{4}A+cos^{4}A)+1=0$

i )

______________________

We know the algebraic identity:

a³+b³+3ab(a+b) = (a+b)³

=> a³+b³ = (a+b)³-3ab(a+b)

By Trigonometric identity:

sin²A + cos²A = 1

______________________

$sin^{6}A+cos^{6}A$

= $(sin^{2})^{3}+(cos^{2})^{3}$

= (sin²A+cos²)³-3sin²Acos²A(sin²Acos²A)

= 1-3sin²Acos²A ----(1)

ii )

______________________

We know the algebraic identity:

a²+b²+2ab = (a+b)²

=> a²+b² = (a+b)²-2ab

______________________

Here ,

sin⁴A + cos⁴A

= (sin²A)²+(cos²A)²

= (sin²A+cos²A)²-2sin²Acos²A

= 1 - 2sin²Acos²A ----(2)

Now ,

LHS =$2(sin^{6}A+cos^{6})-3(sin^{4}A+cos^{4}A)+1$

/* from (1)&(2),we get

= 2(1-3sin²Acos²A)-3(1-2sin²Acos²A)+1

= 2-6sin²Acos²A-3-6sin²Acos²A+1

= 2-3+1

= 0

= RHS

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