What is ${sin}^{6} θ$ $sin^6theta$ in terms of non-exponential trigonometric functions?

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What is ${sin}^{6} θ$ $sin^6theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-07-01T23:24:10

${sin}^{6} θ = \frac{1}{32} (10 - cos (6 θ) + 6 cos (4 θ) - 15 cos (2 θ))$ $sin^6 theta=1/32(10-cos(6 theta)+6cos(4 theta)-15cos(2 theta))$

Explanation:

De Moivre's formula tells us that:

${(cos (x) + i sin (x))}^{n} = cos (n x) + i sin (n x)$ $(cos(x)+i sin(x))^n = cos(nx)+i sin(nx)$

For brevity write, $c$ $c$ for $cos θ$ $cos theta$ and $s$ $s$ for $sin θ$ $sin theta$

Use Pythagoras:

$s^{2} + c^{2} = 1$ $s^2+c^2=1$

So:

$cos (6 θ) + i sin (6 θ)$ $cos(6 theta)+i sin(6 theta)$

$= {(c + i s)}^{6}$ $=(c+is)^6$

$= c^{6} + 6 i c^{5} s - 15 c^{4} s^{2} - 20 i c^{3} s^{3} + 15 c^{2} s^{4} + 6 i c s^{5} - s^{6}$ $=c^6+6i c^5s-15 c^4s^2-20i c^3s^3+15 c^2 s^4+6i cs^5-s^6$

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

Equating Real parts:

$cos (6 θ) = c^{6} - 15 c^{4} s^{2} + 15 c^{2} s^{4} - s^{6}$ $cos(6 theta) = c^6-15c^4s^2+15c^2s^4-s^6$

$= {(1 - s^{2})}^{3} - 15 {(1 - s^{2})}^{2} s^{2} + 15 (1 - s^{2}) s^{4} - s^{6}$ $=(1-s^2)^3-15(1-s^2)^2s^2+15(1-s^2)s^4-s^6$

$= (1 - 3 s^{2} + 3 s^{4} - s^{6}) - 15 (1 - 2 s^{2} + s^{4}) s^{2} + 15 (1 - s^{2}) s^{4} - s^{6}$ $=(1-3s^2+3s^4-s^6)-15(1-2s^2+s^4)s^2+15(1-s^2)s^4-s^6$

$= 1 - 3 s^{2} + 3 s^{4} - s^{6} - 15 s^{2} + 30 s^{4} - 15 s^{6} + 15 s^{4} - 15 s^{6} - s^{6}$ $=1-3s^2+3s^4-s^6-15s^2+30s^4-15s^6+15s^4-15s^6-s^6$

$= 1 - 18 s^{2} + 48 s^{4} - 32 s^{6}$ $=1-18s^2+48s^4-32s^6$

$cos (4 θ) + i sin (4 θ)$ $cos(4 theta)+i sin(4 theta)$

$= {(c + i s)}^{4}$ $=(c+is)^4$

$= c^{4} + 4 i c^{3} s - 6 c^{2} s^{2} - 4 i c s^{3} + s^{4}$ $=c^4+4ic^3s-6c^2s^2-4ics^3+s^4$

$= (c^{4} - 6 c^{2} s^{2} + s^{4}) + i (4 c^{3} s - 4 c s^{3})$ $=(c^4-6c^2s^2+s^4)+i(4c^3s-4cs^3)$

Equating Real parts:

$cos (4 θ) = c^{4} - 6 c^{2} s^{2} + s^{4}$ $cos(4 theta) = c^4-6c^2s^2+s^4$

$= {(1 - s^{2})}^{2} - 6 (1 - s^{2}) s^{2} + s^{4}$ $=(1-s^2)^2-6(1-s^2)s^2+s^4$

$= (1 - 2 s^{2} + s^{4}) - 6 (1 - s^{2}) s^{2} + s^{4}$ $=(1-2s^2+s^4)-6(1-s^2)s^2+s^4$

$= 1 - 2 s^{2} + s^{4} - 6 s^{2} + 6 s^{4} + s^{4}$ $=1-2s^2+s^4-6s^2+6s^4+s^4$

$= 1 - 8 s^{2} + 8 s^{4}$ $=1-8s^2+8s^4$

$cos (2 θ) + i sin (2 θ)$ $cos(2 theta)+i sin(2 theta)$

$= {(c + i s)}^{2}$ $=(c+is)^2$

$= c^{2} + 2 i c s - s^{2}$ $=c^2+2ics-s^2$

$= (c^{2} - s^{2}) + i (2 c s)$ $=(c^2-s^2)+i(2cs)$

Equating Real parts:

$cos (2 θ) = c^{2} - s^{2} = (1 - s^{2}) - s^{2} = 1 - 2 s^{2}$ $cos(2 theta) = c^2-s^2 = (1-s^2)-s^2 = 1-2s^2$

So:

$cos (6 θ) - 6 cos (4 θ)$ $cos(6 theta)-6cos(4 theta)$

$= (1 - 18 s^{2} + 48 s^{4} - 32 s^{6}) - 6 (1 - 8 s^{2} + 8 s^{4})$ $=(1-18s^2+48s^4-32s^6)-6(1-8s^2+8s^4)$

$= 1 - 18 s^{2} + 48 s^{4} - 32 s^{6} - 6 + 48 s^{2} - 48 s^{4}$ $=1-18s^2+48s^4-32s^6-6+48s^2-48s^4$

$= - 5 + 30 s^{2} - 32 s^{6}$ $=-5+30s^2-32s^6$

So:

$cos (6 θ) - 6 cos (4 θ) + 15 cos (2 θ)$ $cos(6 theta)-6cos(4 theta)+15cos(2 theta)$

$= (- 5 + 30 s^{2} - 32 s^{6}) + 15 (1 - 2 s^{2})$ $=(-5+30s^2-32s^6)+15(1-2s^2)$

$= - 5 + 30 s^{2} - 32 s^{6} + 15 - 30 s^{2}$ $=-5+30s^2-32s^6+15-30s^2$

$= 10 - 32 s^{6}$ $=10-32s^6$

Hence:

${sin}^{6} θ = \frac{1}{32} (10 - cos (6 θ) + 6 cos (4 θ) - 15 cos (2 θ))$ $sin^6 theta=1/32(10-cos(6 theta)+6cos(4 theta)-15cos(2 theta))$

Answer 2 · 2016-07-02T01:53:18

$(\frac{1}{8}) (1 - cos 2 t) (1 - cos 2 t) (1 - cos 2 t)$ $(1/8)(1 - cos 2t)(1 - cos 2t)(1 - cos 2t)$

Explanation:

Apply the trig identity: $2 {sin}^{2} t = 1 - cos 2 t$ $2sin^2 t = 1 - cos 2t$
${sin}^{6} t = {sin}^{2} t . {sin}^{2} t . {sin}^{2} t =$ $sin^6 t = sin^2t.sin^2 t.sin^2 t =$
$= (\frac{1 - cos 2 t}{2}) (\frac{1 - cos 2 t}{2}) (\frac{1 - cos 2 t}{2})$ $= ((1 - cos 2t)/2)((1- cos 2t)/2)((1 - cos 2t)/2)$
$= (\frac{1}{8}) (1 - cos 2 t) (1 - cos 2 t) (1 - cos 2 t)$ $= (1/8)(1 - cos 2t)(1 - cos 2t)(1 - cos 2t)$

Answer 3 · 2016-07-02T11:18:50

${sin}^{6} (θ) = \frac{1}{32} (10 - 15 cos (2 θ) + 6 cos (4 θ) - cos (6 θ))$ $sin^6(theta)=1/32(10-15cos(2theta)+6cos(4theta)-cos(6theta))$

Explanation:

${sin}^{6} (θ)$ $sin^6(theta)$ is an even function so its expansion must be of the form

${sin}^{6} (θ) = k = 6 \sum k = 0 c_{k} cos (k θ)$ $sin^6(theta) = sum_{k=0}^{k=6}c_k cos(k theta)$

but

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

then

${sin}^{6} θ = c_{0} + c_{2} cos 2 θ + c_{4} cos 4 θ + c_{6} cos 6 θ$ $sin^6 theta = c_0+c_2 cos2theta+c_4cos4theta+c_6cos6theta$

choosing now $θ_{j} = \frac{j 2 π}{4}$ $theta_j =( j2 pi)/4$ for $j = 0, 1, 2, 3$ $j = 0,1,2,3$
and solving the set of linear equations

${sin}^{6} (θ_{j}) = c_{0} + c_{2} cos 2 θ_{j} + c_{4} cos 4 θ_{j} + c_{6} cos 6 θ_{j}$ $sin^6(theta_j) = c_0+c_2 cos2theta_j+c_4cos4theta_j+c_6cos6theta_j$ for $j = 0, 1, 2, 3$ $j = 0,1,2,3$

for $c_{0}, c_{2}, c_{4}, c_{6}$ $c_0,c_2,c_4,c_6$ we obtain

$c_{0} = \frac{5}{16}, c_{2} = - \frac{15}{32}, c_{4} = \frac{3}{16}, c_{6} = - \frac{1}{32}$ $c_0=5/16,c_2=-15/32,c_4=3/16,c_6=-1/32$

so

${sin}^{6} (θ) = \frac{1}{32} (10 - 15 cos (2 θ) + 6 cos (4 θ) - cos (6 θ))$ $sin^6(theta)=1/32(10-15cos(2theta)+6cos(4theta)-cos(6theta))$

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