give the expansion sin^8@ as a series of cos@
plz help me show working
where @ means theta
sin^2(theta) = 1-cos^2(theta)
Therefore:
sin^8(theta) = [1-cos^2(theta)]^4 =
1 - 4 cos^2(theta) + 6 cos^4(theta) - 4 cos^6(theta) + cos^8(theta)
plz help me show working
where @ means theta
Therefore:
sin^8(theta) = [1-cos^2(theta)]^4 =
1 - 4 cos^2(theta) + 6 cos^4(theta) - 4 cos^6(theta) + cos^8(theta)