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

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Answer 1 · 2018-02-22T13:59:40

See the steps below!

#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!

Answer 2 · 2018-02-22T14:14:56

Please refer to a Proof given in the 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!