Question

Evaluate sin^6-cos^6

Answer 1

Sin^6∅ - cos^6∅
= (sin²∅)³ - (cos²∅)³
use the formula,
a³ - b³ = (a - b)(a² + b² + ab )

= (sin²∅ - cos²∅)(sin⁴∅+ cos⁴∅+ sin²∅.cos²∅)
= -(cos²∅- sin²∅){(sin²∅ + cos²∅)² - 2sin²∅.cos²∅ + sin²∅.cos²∅}
= - cos2∅.{ 1 - 1/4(2sin∅.cos∅)²}
= -cos2∅(4 - sin²2∅)/4

Answer 2

sin^6-cos^6

=(sin^2)^3-(cos^2)^3

{a^3-b^3=(a-b)(a^2+b^2+ab)

=(sin^2-cos^2)(sin^4+sin^2cos^2+cos^4)

{a^2+b^2+ab=(a+b)^2 -ab}

=(sin^2-cos^2){(sin^2+cos^2)^2-sin^2cos^2}

identity (sin^2+cos^2)=1

=(sin^2-cos^2)(1-sin^2cos^2)

=(sin^2-cos^2){sin^2+cos^2-(1-sin^2)cos^2}

=(sin^2-cos^2)(sin^4+cos^2)

#lordcarbin.