What is #sin^6theta# in terms of non-exponential trigonometric functions?

3 Answers
George C.
Jul 1, 2016

#sin^6 theta=1/32(10-cos(6 theta)+6cos(4 theta)-15cos(2 theta))#

Explanation:

De Moivre's formula tells us that:

#(cos(x)+i sin(x))^n = cos(nx)+i sin(nx)#

For brevity write, #c# for #cos theta# and #s# for #sin theta#

Use Pythagoras:

#s^2+c^2=1#

So:

#cos(6 theta)+i sin(6 theta)#

#=(c+is)^6#

#=c^6+6i c^5s-15 c^4s^2-20i c^3s^3+15 c^2 s^4+6i cs^5-s^6#

#=(c^6-15c^4s^2+15c^2s^4-s^6)+i(6c^5s-20c^3s^3+6cs^5)#

Equating Real parts:

#cos(6 theta) = c^6-15c^4s^2+15c^2s^4-s^6#

#=(1-s^2)^3-15(1-s^2)^2s^2+15(1-s^2)s^4-s^6#

#=(1-3s^2+3s^4-s^6)-15(1-2s^2+s^4)s^2+15(1-s^2)s^4-s^6#

#=1-3s^2+3s^4-s^6-15s^2+30s^4-15s^6+15s^4-15s^6-s^6#

#=1-18s^2+48s^4-32s^6#

#cos(4 theta)+i sin(4 theta)#

#=(c+is)^4#

#=c^4+4ic^3s-6c^2s^2-4ics^3+s^4#

#=(c^4-6c^2s^2+s^4)+i(4c^3s-4cs^3)#

Equating Real parts:

#cos(4 theta) = c^4-6c^2s^2+s^4#

#=(1-s^2)^2-6(1-s^2)s^2+s^4#

#=(1-2s^2+s^4)-6(1-s^2)s^2+s^4#

#=1-2s^2+s^4-6s^2+6s^4+s^4#

#=1-8s^2+8s^4#

#cos(2 theta)+i sin(2 theta)#

#=(c+is)^2#

#=c^2+2ics-s^2#

#=(c^2-s^2)+i(2cs)#

Equating Real parts:

#cos(2 theta) = c^2-s^2 = (1-s^2)-s^2 = 1-2s^2#

So:

#cos(6 theta)-6cos(4 theta)#

#=(1-18s^2+48s^4-32s^6)-6(1-8s^2+8s^4)#

#=1-18s^2+48s^4-32s^6-6+48s^2-48s^4#

#=-5+30s^2-32s^6#

So:

#cos(6 theta)-6cos(4 theta)+15cos(2 theta)#

#=(-5+30s^2-32s^6)+15(1-2s^2)#

#=-5+30s^2-32s^6+15-30s^2#

#=10-32s^6#

Hence:

#sin^6 theta=1/32(10-cos(6 theta)+6cos(4 theta)-15cos(2 theta))#

Nghi N.
Jul 2, 2016

#(1/8)(1 - cos 2t)(1 - cos 2t)(1 - cos 2t)#

Explanation:

Apply the trig identity: #2sin^2 t = 1 - cos 2t#
#sin^6 t = sin^2t.sin^2 t.sin^2 t =#
#= ((1 - cos 2t)/2)((1- cos 2t)/2)((1 - cos 2t)/2)#
#= (1/8)(1 - cos 2t)(1 - cos 2t)(1 - cos 2t)#

Cesareo R.
Jul 2, 2016

#sin^6(theta)=1/32(10-15cos(2theta)+6cos(4theta)-cos(6theta))#

Explanation:

#sin^6(theta)# is an even function so its expansion must be of the form

#sin^6(theta) = sum_{k=0}^{k=6}c_k cos(k theta)#

but

#int_{-pi}^{pi}sin^6 theta cos((2k+1)theta) d theta = 0# for #k = 0,1,2#

then

#sin^6 theta = c_0+c_2 cos2theta+c_4cos4theta+c_6cos6theta#

choosing now #theta_j =( j2 pi)/4# for #j = 0,1,2,3#
and solving the set of linear equations

#sin^6(theta_j) = c_0+c_2 cos2theta_j+c_4cos4theta_j+c_6cos6theta_j# for #j = 0,1,2,3#

for #c_0,c_2,c_4,c_6# we obtain

#c_0=5/16,c_2=-15/32,c_4=3/16,c_6=-1/32#

so

#sin^6(theta)=1/32(10-15cos(2theta)+6cos(4theta)-cos(6theta))#

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