Prove that 2(sin^6a+cos^6a)-3(sin ^4+cos^4a)+1=?

2 Answers
Feb 22, 2018

See the steps below!

Explanation:

2(sin^6a+cos^6a)-3(sin^4a+cos^4a)+1=0

Rewrite the expression by applying the distributive rule a^3+b^3=(a+b)(a^2-ab+b^2)

2(sin^2a+cos^2a)(sin^4a-sin^2acos^2b+cos^2a)-3(sin^4a+cos^4a)+1=0

Put the value of pythagorean identity \ \ \ sin^2(x)+cos^2(x)=1, to get:

2(1)(sin^4a-sin^2acos^2b+cos^2a)-3(sin^4a+cos^4a)+1=0

2sin^4a-2sin^2acos^2b+2cos^4a-3sin^4a-3cos^4a+1=0

Simplify:

-sin^4(a)-2sin^2acos^2b-cos^4a+1=0

-(sin^4a+2sin^2acos^2b+cos^4a)+1=0

-(sin^2a+cos^2a)^2+1=0

Put the value of pythagorean identity \ \ \ sin^2(x)+cos^2(x)=1, to get:

-1+1=0

0=0

That's it!

Feb 22, 2018

Please refer to a Proof given in the Explanation.

Explanation:

Recall that, x+y+z=0rArr x^3+y^3+z^3=3xyz.

Using this Result, we have,

sin^2A+cos^2A+(-1)=0,

rArr(sin^2A)^3+(cos^2A)^3+(-1)^3=3(sin^2A)(cos^2A)(-1)

:.sin^6A+cos^6A=1-3sin^2Acos^2A.....................(diamond1).

Also, (sin^2A+cos^2A)=1 rArr (sin^2A+cos^2A)^2=1^2.

:. sin^4A+2sin^2Acos^2A+cos^4A=1, or,

sin^4A+cos^4A=1-2sin^2Acos^2A........................(diamond2).

Hence, utilising (diamond1) and (diamond2), we get,

2(sin^6A+cos^6A)-3(sin^4A+cos^4A),

=2(1-3sin^2Acos^2A)-3(1-2sin^2Acos^2A),

=-1.

rArr 2(sin^6A+cos^6A)-3(sin^4A+cos^4A)+1=0,

as desired!