Question

Evaluate : Sin^6x - Cos^6x

Answer 1

sin6x−cos6xsin6⁡x−cos6⁡x

=(sin2x)3−(cos2x)3=(sin2⁡x)3−(cos2⁡x)3

=(sin2x−cos2x)(sin4x+sin2xcos2x+cos4x)=(sin2⁡x−cos2⁡x)(sin4⁡x+sin2⁡xcos2⁡x+cos4⁡x)

=−cos2x[(sin2x)2+(cos2x)2+sin2xcos2x]=−cos⁡2x[(sin2⁡x)2+(cos2⁡x)2+sin2⁡xcos2⁡x]

=−cos2x[(sin2x+cos2x)2−2sin2xcos2x+sin2xcos2x]=−cos⁡2x[(sin2⁡x+cos2⁡x)2−2sin2⁡xcos2⁡x+sin2⁡xcos2⁡x]

=−cos2x(1−sin2xcos2x)=−cos⁡2x(1−sin2⁡xcos2⁡x)

=−cos2x(1−sin22x4)=−cos⁡2x(1−sin2⁡2x4)

=sin2xsin4x8−cos2x

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Answer 2

Sin⁶x  - Cos⁶x = (Sin²x  - Cos²x)(1 - Sin²xCos²x) = (Sinx + Cosx)(Sinx - Cosx)(1 + SinxCosx)(1 - SinxCosx)

Step-by-step explanation:

Sin⁶x  - Cos⁶x

= (Sin³x)² - (Cos³x)²

using a² - b² = (a + b)(a - b)

= (Sin³x + Cos³x)(Sin³x - Cos³x)

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

= (Sinx + Cosx)(Sin²x - SinxCosx + Cos²x) (Sinx - Cosx)(Sin²x + SinxCosx + Cos²x)

using Sin²x  + Cos²x = 1

= (Sinx + Cosx)(1 - SinxCosx) (Sinx - Cosx)(1 + SinxCosx)

= (Sin²x  - Cos²x)(1 - Sin²xCos²x)

Sin⁶x  - Cos⁶x = (Sin²x  - Cos²x)(1 - Sin²xCos²x)

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