Question

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

Answer 1

Hint: We use some concepts of complex numbers and their properties to solve this problem. We use some algebraic identities like
\[{(a + b)^6} = {a^6} + 6{a^5}b + 15{a^4}{b^2} + 20{a^3}{b^3} + 15{a^2}{b^4} + 6a{b^5} + {b^6}\]
\[{(a + b)^4} = {a^4} + 4{a^3}b + 6{a^2}{b^2} + 4a{b^3} + {b^4}\]

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 \[a + ib\] where \[a\] is the real part and \[b\] is the imaginary part. Some examples of complex numbers are \[1, - \dfrac{5}{7},(3 + 4i)\] and so on.
We all know the De-Moivre’s theorem which tells us that, \[{\left( {\cos (\theta ) + i\sin (\theta )} \right)^n} = \cos (n\theta ) + i\sin (n\theta )\]
Here, let the value of \[n\] be 6.
So, \[{\left( {\cos (\theta ) + i\sin (\theta )} \right)^6} = \cos (6\theta ) + i\sin (6\theta )\]
Let \[\cos \theta = c\] and \[\sin \theta = s\] for our convenience. So, that implies \[{c^2} + {s^2} = 1\] -----(1)
So, now let’s evaluate the left-hand part, which is \[{\left( {\cos (\theta ) + i\sin (\theta )} \right)^6} = {(c + is)^6}\]
So, \[{(c + is)^6} = {c^6} + 6i{c^5}s - 15{c^4}{s^2} - 20i{c^3}{s^3} + 15{c^2}{s^4} + 6ic{s^5} - {s^6}\]
\[ \Rightarrow {(c + is)^6} = ({c^6} - 15{c^4}{s^2} + 15{c^2}{s^4} - {s^6}) + i(6{c^5}s - 20{c^3}{s^3} + 6c{s^5})\]
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 \[\cos (6\theta ) = {c^6} - 15{c^4}{s^2} + 15{c^2}{s^4} - {s^6}\]
Now, from (1), we get, \[{c^2} = 1 - {s^2}\] and substitute this in above.
\[ \Rightarrow \cos (6\theta ) = {\left( {1 - {s^2}} \right)^3} - 15{\left( {1 - {s^2}} \right)^2}{s^2} + 15\left( {1 - {s^2}} \right){s^4} - {s^6}\]
\[ \Rightarrow \cos (6\theta ) = \left( {1 - 3{s^2} + 3{s^4} - {s^6}} \right) - 15\left( {1 - 2{s^2} + {s^4}} \right){s^2} + 15\left( {1 - {s^2}} \right){s^4} - {s^6}\]
Now, let us simplify further.
\[ \Rightarrow \cos (6\theta ) = 1 - 3{s^2} + 3{s^4} - {s^6} - \left( {15{s^2} - 30{s^4} + 15{s^6}} \right) + \left( {15{s^4} - 15{s^6}} \right) - {s^6}\]
\[ \Rightarrow \cos (6\theta ) = 1 - 18{s^2} + 48{s^4} - 32{s^6}\] ------(2)
Now in the same way, we will evaluate \[\cos (4\theta ) + i\sin (4\theta ) = {\left( {c + is} \right)^4}\]
\[ \Rightarrow {\left( {c + is} \right)^4} = {c^4} + 4i{c^3}s - 6{c^2}{s^2} - 4ic{s^3} + {s^4}\]
\[ \Rightarrow \cos (4\theta ) + i\sin (4\theta ) = ({c^4} - 6{c^2}{s^2} + {s^4}) + i(4{c^3}s - 4c{s^3})\]
Equating the real parts, we get,
\[\cos (4\theta ) = {c^4} - 6{c^2}{s^2} + {s^4}\]
\[ \Rightarrow \cos (4\theta ) = {(1 - {s^2})^2} - 6(1 - {s^2}){s^2} + {s^4}\]
So, we get,
\[ \Rightarrow \cos (4\theta ) = (1 - 2{s^2} + {s^4}) - 6({s^2} - {s^4}) + {s^4}\]
\[ \Rightarrow \cos (4\theta ) = 1 - 8{s^2} + 8{s^4}\] ------(3)
Now let us evaluate \[\cos (2\theta ) + i\sin (2\theta ) = {(c + is)^2}\]
\[ \Rightarrow {(c + is)^2} = ({c^2} - {s^2}) + i(2cs)\]
Equating real parts,
\[ \Rightarrow \cos (2\theta ) = {c^2} - {s^2} = (1 - {s^2}) - {s^2}\]
\[ \Rightarrow \cos (2\theta ) = 1 - 2{s^2}\] -----(4)
Now, derive the value of \[{s^2}\] from this equation.
\[{s^2} = \dfrac{{1 - \cos (2\theta )}}{2}\]
Now, substitute this value in equation (3), and find the value of \[{s^4}\]
\[ \Rightarrow \cos (4\theta ) = 1 - 8\left( {\dfrac{{1 - \cos (2\theta )}}{2}} \right) + 8{s^4}\]
\[ \Rightarrow {s^4} = \dfrac{{\cos (4\theta ) + 3 - 4\cos (2\theta )}}{8}\]
Now, substitute this value in equation (2) and find the value of \[{s^6}\]
\[ \Rightarrow \cos (6\theta ) = 1 - 18\left( {\dfrac{{1 - \cos (2\theta )}}{2}} \right) + 48\left( {\dfrac{{\cos (4\theta ) + 3 - 4\cos (2\theta )}}{8}} \right) - 32{s^6}\]
\[ \Rightarrow \cos (6\theta ) = 1 - 9\left( {1 - \cos (2\theta )} \right) + 6\left( {\cos (4\theta ) + 3 - 4\cos (2\theta )} \right) - 32{s^6}\]
So, we get it as,
\[ \Rightarrow \cos (6\theta ) = 1 - 9 + 9\cos (2\theta ) + 6\cos (4\theta ) + 18 - 24\cos (2\theta ) - 32{s^6}\]
\[ \Rightarrow \cos (6\theta ) = 6\cos (4\theta ) + 10 - 15\cos (2\theta ) - 32{s^6}\]
So, we finally get,
\[ \Rightarrow {s^6} = \dfrac{1}{{32}}\left( {6\cos (4\theta ) + 10 - 15\cos (2\theta ) - \cos (6\theta )} \right)\]
And we know that, \[{s^6} = {\sin ^6}\theta \]
So, we can conclude that,
\[{\sin ^6}\theta = \dfrac{1}{{32}}\left( {6\cos (4\theta ) - 15\cos (2\theta ) - \cos (6\theta ) + 10} \right)\]

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