Question

The expansion of \[{\sin ^8}\theta \] will be the series of
A. sine multiple
B. cosine multiple
C. both sine and cosine multiple
D. can’t be determined

Answer 1

Hint: Here in this question we have determined the \[{\sin ^8}\theta \]. On expanding the term we have to choose the options. Firstly we consider \[x = \cos \theta + i\sin \theta \], \[\dfrac{1}{x} = \cos \theta - i\sin \theta \], \[{x^n} = \cos n\theta + i\sin n\theta \] and \[\dfrac{1}{{{x^n}}} = \cos n\theta - i\sin n\theta \]. Then on expanding the term raised to the power 8 and simplification we obtain the required solution for the given question.

Complete step by step answer:
As we know that \[x = \cos \theta + i\sin \theta \] ---- (1) and \[\dfrac{1}{x} = \cos \theta - i\sin \theta \]----(2)
On subtracting equation (1) and equation (2) we have
\[ \Rightarrow x - \dfrac{1}{x} = \cos \theta + i\sin \theta - \cos \theta + i\sin \theta \]
On simplifying we have
\[ \Rightarrow x - \dfrac{1}{x} = 2i\sin \theta \]---- (3)
On adding the equation (1) and equation (2)
\[ \Rightarrow x + \dfrac{1}{x} = \cos \theta + i\sin \theta + \cos \theta - i\sin \theta \]
On simplifying we have
\[ \Rightarrow x + \dfrac{1}{x} = 2\cos \theta \]---- (4)
As we know that \[{x^n} = \cos n\theta + i\sin n\theta \] ---- (5) and \[\dfrac{1}{{{x^n}}} = \cos n\theta - i\sin n\theta \]----(6)
On subtracting equation (5) and equation (6) we have
\[ \Rightarrow {x^n} - \dfrac{1}{{{x^n}}} = \cos n\theta + i\sin n\theta - \cos n\theta + i\sin n\theta \]

On simplifying we have
\[ \Rightarrow {x^n} - \dfrac{1}{{{x^n}}} = 2i\sin n\theta \]---- (7)
On adding the equation (5) and equation (6)
\[ \Rightarrow {x^n} + \dfrac{1}{{{x^n}}} = \cos n\theta + i\sin n\theta + \cos n\theta - i\sin n\theta \]
On simplifying we have
\[ \Rightarrow {x^n} + \dfrac{1}{{{x^n}}} = 2\cos n\theta \]---- (8)
Now we have to find the expansion of \[{\sin ^8}\theta \].
On considering the equation (3) and raised to the power of 8.
\[ \Rightarrow {\left( {x - \dfrac{1}{x}} \right)^8} = {\left( {2i\sin \theta } \right)^8}\] ------(9)
The expansion \[{(a - b)^8} = {a^8} - 8{a^7}.b + 28{a^6}.{b^2} - 56{a^5}.{b^3} + 70{a^4}.{b^4} - 56{a^3}.{b^5} + 28{a^2}.{b^6} - 8a.{b^7} + {b^8}\]
So the equation (9) can be written as
\[ \Rightarrow {x^8} - 8{x^7}.\dfrac{1}{x} + 28{x^6}.\dfrac{1}{{{x^2}}} - 56{x^5}.\dfrac{1}{{{x^3}}} + 70{x^4}.\dfrac{1}{{{x^4}}} - 56{x^3}.\dfrac{1}{{{x^5}}} + 28{x^2}.\dfrac{1}{{{x^6}}} - 8x.\dfrac{1}{{{x^7}}} + \dfrac{1}{{{x^8}}} = 256\,{\sin ^8}\theta \]

On simplifying this we have
\[ \Rightarrow {x^8} - 8{x^6} + 28{x^4} - 56{x^2} + 70 - 56\dfrac{1}{{{x^2}}} + 28\dfrac{1}{{{x^4}}} - 8\dfrac{1}{{{x^6}}} + \dfrac{1}{{{x^8}}} = 256\,{\sin ^8}\theta \]
Taking the common terms in the LHS we have
\[ \Rightarrow \left( {{x^8} + \dfrac{1}{{{x^8}}}} \right) - 8\left( {{x^6} + \dfrac{1}{{{x^6}}}} \right) + 28\left( {{x^4} + \dfrac{1}{{{x^4}}}} \right) - 56\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) + 70 = 256\,{\sin ^8}\theta \]
On considering the equation (8), the term is written as
\[ \Rightarrow \left( {2\cos 8\theta } \right) - 8\left( {2\cos 6\theta } \right) + 28\left( {2\cos 4\theta } \right) - 56\left( {2\cos 2\theta } \right) + 70 = 256\,{\sin ^8}\theta \]
On simplifying we have
\[ \Rightarrow 2\cos 8\theta - 16\cos 6\theta + 56\cos 4\theta - 112\cos 2\theta + 70 = 256\,{\sin ^8}\theta \]

On dividing the above term by 2 we get
\[ \Rightarrow \cos 8\theta - 8\cos 6\theta + 28\cos 4\theta - 56\cos 2\theta + 35 = 128\,{\sin ^8}\theta \]
This can be written as
\[ \Rightarrow {\sin ^8}\theta = \dfrac{1}{{128}}\left[ {\cos 8\theta - 8\cos 6\theta + 28\cos 4\theta - 56\cos 2\theta + 35} \right]\]
As we know that the
\[\cos n\theta = {\cos ^n}\theta - \left( {\begin{array}{*{20}{c}}
  n \\
  2
\end{array}} \right){\cos ^{n - 2}}\theta \,{\sin ^2}\theta + \left( {\begin{array}{*{20}{c}}
  n \\
  4
\end{array}} \right){\cos ^{n - 4}}\theta \,{\sin ^4}\theta \pm ...\], on considering this formula the \[{\sin ^8}\theta \] will be the series of sine and cosine.

Therefore the correct answer is option C.

Note:We should note that pascal’s triangle is helpful only when the value of $n$ is small in the equation ${(a - b)^n}$. If the value is large then it is very tedious to draw the triangle until we reach $n$. Between the two terms we have negative so we will have alternate signs in the expansion.