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

Jul 1, 2016

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

#### Explanation:

De Moivre's formula tells us that:

${\left(\cos \left(x\right) + i \sin \left(x\right)\right)}^{n} = \cos \left(n x\right) + i \sin \left(n x\right)$

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

Use Pythagoras:

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

So:

$\cos \left(6 \theta\right) + i \sin \left(6 \theta\right)$

$= {\left(c + i s\right)}^{6}$

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

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

Equating Real parts:

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

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

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

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

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

$\cos \left(4 \theta\right) + i \sin \left(4 \theta\right)$

$= {\left(c + i s\right)}^{4}$

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

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

Equating Real parts:

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

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

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

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

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

$\cos \left(2 \theta\right) + i \sin \left(2 \theta\right)$

$= {\left(c + i s\right)}^{2}$

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

$= \left({c}^{2} - {s}^{2}\right) + i \left(2 c s\right)$

Equating Real parts:

$\cos \left(2 \theta\right) = {c}^{2} - {s}^{2} = \left(1 - {s}^{2}\right) - {s}^{2} = 1 - 2 {s}^{2}$

So:

$\cos \left(6 \theta\right) - 6 \cos \left(4 \theta\right)$

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

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

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

So:

$\cos \left(6 \theta\right) - 6 \cos \left(4 \theta\right) + 15 \cos \left(2 \theta\right)$

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

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

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

Hence:

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

Jul 2, 2016

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

#### Explanation:

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

Jul 2, 2016

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

#### Explanation:

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

${\sin}^{6} \left(\theta\right) = {\sum}_{k = 0}^{k = 6} {c}_{k} \cos \left(k \theta\right)$

but

${\int}_{- \pi}^{\pi} {\sin}^{6} \theta \cos \left(\left(2 k + 1\right) \theta\right) d \theta = 0$ for $k = 0 , 1 , 2$

then

${\sin}^{6} \theta = {c}_{0} + {c}_{2} \cos 2 \theta + {c}_{4} \cos 4 \theta + {c}_{6} \cos 6 \theta$

choosing now ${\theta}_{j} = \frac{j 2 \pi}{4}$ for $j = 0 , 1 , 2 , 3$
and solving the set of linear equations

${\sin}^{6} \left({\theta}_{j}\right) = {c}_{0} + {c}_{2} \cos 2 {\theta}_{j} + {c}_{4} \cos 4 {\theta}_{j} + {c}_{6} \cos 6 {\theta}_{j}$ for $j = 0 , 1 , 2 , 3$

for ${c}_{0} , {c}_{2} , {c}_{4} , {c}_{6}$ we obtain

${c}_{0} = \frac{5}{16} , {c}_{2} = - \frac{15}{32} , {c}_{4} = \frac{3}{16} , {c}_{6} = - \frac{1}{32}$

so

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