Question

evaluate cos ^6 thita +sin^6 thita​

Answer 1

Step-by-step explanation:

zaywqwemdk have a great day to

Answer 2

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

[ use, cos²x - sin²x = cos2x , 2sinx.cosx = sin2x ]

hope it helps...

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