evaluate sin raise to 6 theta - cos raise to 6 theta
Dhinu:
is 6 the power of sine ??
pr is it coefficient of theta ??
or*
fine..... :)
i can't give u the answer of both ...
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Answered by
248
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 ]
= (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 ]
thanx ☺
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17
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