Sin(4x) - sin(8x)
sin(2(2x)) - sin(2(4x))
2sin(2x)cos(2x) - 2sin(4x)cos(4x)
2[sin(x)cos(x)][cos²(x) - sin²(x)] - [2sin(2(2x))cos(2(2x))]
{2[sin(x)cos³(x) - sin³(x)cos(x)]} - [2[2sin(2x)cos(2x)][cos²(x) - sin²(x)]]
[2sin(x)cos³(x) - 2sin³(x)cos(x)] - {[2[2[sin(x)cos(x)][cos²(x) - sin²(x)]][cos²(x) - sin²(x)]}
[2sin(x)cos³(x) - 2sin³(x)cos(x)] - [2[2[sin(x)cos(x)][cos⁴(x) - 2cos²(x)sin²(x) - sin⁴(x)]]]]
[2sin(x)cos³(x) - 2sin³(x)cos(x)] - [2[2[sin(x)cos⁵(x) - 2sin³(x)cos³(x) + sin⁵(x)cos(x)]]]
2sin(x)cos³(x) - 2sin³(x)cos(x) - 4sin(x)cos⁵(x) + 8sin³(x)cos³(x) - 4sin⁵(x)cos(x)
Answer:
A
Step-by-step explanation:
Make them have the same denominator.
1/2 + 1/3
=3/6 + 2/6
=5/6
And what else is in the problem ??????
3bars/4boys=0.75bars per boy. or 3/4th bars per boy.
