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Question: What is \[{\sin ^6}\theta \] in terms of non-exponential trigonometric function?...

What is sin6θ{\sin ^6}\theta in terms of non-exponential trigonometric function?

Explanation

Solution

We use some concepts of complex numbers and their properties to solve this problem. We use some algebraic identities like
(a+b)6=a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6{(a + b)^6} = {a^6} + 6{a^5}b + 15{a^4}{b^2} + 20{a^3}{b^3} + 15{a^2}{b^4} + 6a{b^5} + {b^6}
(a+b)4=a4+4a3b+6a2b2+4ab3+b4{(a + b)^4} = {a^4} + 4{a^3}b + 6{a^2}{b^2} + 4a{b^3} + {b^4}

Complete step by step answer:
A complex number is an imaginary number which has no definite value. All the rational and irrational numbers come under these complex numbers. It is written as a+iba + ib where aa is the real part and bb is the imaginary part. Some examples of complex numbers are 1,57,(3+4i)1, - \dfrac{5}{7},(3 + 4i) and so on.
We all know the De-Moivre’s theorem which tells us that, (cos(θ)+isin(θ))n=cos(nθ)+isin(nθ){\left( {\cos (\theta ) + i\sin (\theta )} \right)^n} = \cos (n\theta ) + i\sin (n\theta )
Here, let the value of nn be 6.
So, (cos(θ)+isin(θ))6=cos(6θ)+isin(6θ){\left( {\cos (\theta ) + i\sin (\theta )} \right)^6} = \cos (6\theta ) + i\sin (6\theta )
Let cosθ=c\cos \theta = c and sinθ=s\sin \theta = s for our convenience. So, that implies c2+s2=1{c^2} + {s^2} = 1 -----(1)
So, now let’s evaluate the left-hand part, which is (cos(θ)+isin(θ))6=(c+is)6{\left( {\cos (\theta ) + i\sin (\theta )} \right)^6} = {(c + is)^6}
So, (c+is)6=c6+6ic5s15c4s220ic3s3+15c2s4+6ics5s6{(c + is)^6} = {c^6} + 6i{c^5}s - 15{c^4}{s^2} - 20i{c^3}{s^3} + 15{c^2}{s^4} + 6ic{s^5} - {s^6}
(c+is)6=(c615c4s2+15c2s4s6)+i(6c5s20c3s3+6cs5)\Rightarrow {(c + is)^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} + 6c{s^5})
So, we can conclude that,
{\left( {\cos (\theta ) + i\sin (\theta )} \right)^6} = \cos (6\theta ) + i\sin (6\theta )$$$$ = {(c + is)^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} + 6c{s^5})
Now, equating the real parts, we get cos(6θ)=c615c4s2+15c2s4s6\cos (6\theta ) = {c^6} - 15{c^4}{s^2} + 15{c^2}{s^4} - {s^6}
Now, from (1), we get, c2=1s2{c^2} = 1 - {s^2} and substitute this in above.
cos(6θ)=(1s2)315(1s2)2s2+15(1s2)s4s6\Rightarrow \cos (6\theta ) = {\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}
cos(6θ)=(13s2+3s4s6)15(12s2+s4)s2+15(1s2)s4s6\Rightarrow \cos (6\theta ) = \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}
Now, let us simplify further.
cos(6θ)=13s2+3s4s6(15s230s4+15s6)+(15s415s6)s6\Rightarrow \cos (6\theta ) = 1 - 3{s^2} + 3{s^4} - {s^6} - \left( {15{s^2} - 30{s^4} + 15{s^6}} \right) + \left( {15{s^4} - 15{s^6}} \right) - {s^6}
cos(6θ)=118s2+48s432s6\Rightarrow \cos (6\theta ) = 1 - 18{s^2} + 48{s^4} - 32{s^6} ------(2)
Now in the same way, we will evaluate cos(4θ)+isin(4θ)=(c+is)4\cos (4\theta ) + i\sin (4\theta ) = {\left( {c + is} \right)^4}
(c+is)4=c4+4ic3s6c2s24ics3+s4\Rightarrow {\left( {c + is} \right)^4} = {c^4} + 4i{c^3}s - 6{c^2}{s^2} - 4ic{s^3} + {s^4}
cos(4θ)+isin(4θ)=(c46c2s2+s4)+i(4c3s4cs3)\Rightarrow \cos (4\theta ) + i\sin (4\theta ) = ({c^4} - 6{c^2}{s^2} + {s^4}) + i(4{c^3}s - 4c{s^3})
Equating the real parts, we get,
cos(4θ)=c46c2s2+s4\cos (4\theta ) = {c^4} - 6{c^2}{s^2} + {s^4}
cos(4θ)=(1s2)26(1s2)s2+s4\Rightarrow \cos (4\theta ) = {(1 - {s^2})^2} - 6(1 - {s^2}){s^2} + {s^4}
So, we get,
cos(4θ)=(12s2+s4)6(s2s4)+s4\Rightarrow \cos (4\theta ) = (1 - 2{s^2} + {s^4}) - 6({s^2} - {s^4}) + {s^4}
cos(4θ)=18s2+8s4\Rightarrow \cos (4\theta ) = 1 - 8{s^2} + 8{s^4} ------(3)
Now let us evaluate cos(2θ)+isin(2θ)=(c+is)2\cos (2\theta ) + i\sin (2\theta ) = {(c + is)^2}
(c+is)2=(c2s2)+i(2cs)\Rightarrow {(c + is)^2} = ({c^2} - {s^2}) + i(2cs)
Equating real parts,
cos(2θ)=c2s2=(1s2)s2\Rightarrow \cos (2\theta ) = {c^2} - {s^2} = (1 - {s^2}) - {s^2}
cos(2θ)=12s2\Rightarrow \cos (2\theta ) = 1 - 2{s^2} -----(4)
Now, derive the value of s2{s^2} from this equation.
s2=1cos(2θ)2{s^2} = \dfrac{{1 - \cos (2\theta )}}{2}
Now, substitute this value in equation (3), and find the value of s4{s^4}
cos(4θ)=18(1cos(2θ)2)+8s4\Rightarrow \cos (4\theta ) = 1 - 8\left( {\dfrac{{1 - \cos (2\theta )}}{2}} \right) + 8{s^4}
s4=cos(4θ)+34cos(2θ)8\Rightarrow {s^4} = \dfrac{{\cos (4\theta ) + 3 - 4\cos (2\theta )}}{8}
Now, substitute this value in equation (2) and find the value of s6{s^6}
cos(6θ)=118(1cos(2θ)2)+48(cos(4θ)+34cos(2θ)8)32s6\Rightarrow \cos (6\theta ) = 1 - 18\left( {\dfrac{{1 - \cos (2\theta )}}{2}} \right) + 48\left( {\dfrac{{\cos (4\theta ) + 3 - 4\cos (2\theta )}}{8}} \right) - 32{s^6}
cos(6θ)=19(1cos(2θ))+6(cos(4θ)+34cos(2θ))32s6\Rightarrow \cos (6\theta ) = 1 - 9\left( {1 - \cos (2\theta )} \right) + 6\left( {\cos (4\theta ) + 3 - 4\cos (2\theta )} \right) - 32{s^6}
So, we get it as,
cos(6θ)=19+9cos(2θ)+6cos(4θ)+1824cos(2θ)32s6\Rightarrow \cos (6\theta ) = 1 - 9 + 9\cos (2\theta ) + 6\cos (4\theta ) + 18 - 24\cos (2\theta ) - 32{s^6}
cos(6θ)=6cos(4θ)+1015cos(2θ)32s6\Rightarrow \cos (6\theta ) = 6\cos (4\theta ) + 10 - 15\cos (2\theta ) - 32{s^6}
So, we finally get,
s6=132(6cos(4θ)+1015cos(2θ)cos(6θ))\Rightarrow {s^6} = \dfrac{1}{{32}}\left( {6\cos (4\theta ) + 10 - 15\cos (2\theta ) - \cos (6\theta )} \right)
And we know that, s6=sin6θ{s^6} = {\sin ^6}\theta
So, we can conclude that,
sin6θ=132(6cos(4θ)15cos(2θ)cos(6θ)+10){\sin ^6}\theta = \dfrac{1}{{32}}\left( {6\cos (4\theta ) - 15\cos (2\theta ) - \cos (6\theta ) + 10} \right)

Note:
Here, ii is a complex value, and it is i=1i = \sqrt { - 1} . So, it can be written as i2=1{i^2} = - 1.
Consider the value in{i^n}. If nn is a multiple of 4, then in=1{i^n} = 1
If nn is an even number other than multiples of 4, then in=1{i^n} = - 1.
An imaginary number is written as a+iba + ib where aa is the real part and bb is the imaginary part.