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

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

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.