Answer

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**Hint:**First use the law of indices for brackets to change the sine function into cosine using their relation identity. Then after converting to cosine, use the algebraic identity ${(a - b)^2} = {a^2} - 2ab + {b^2}$ in order to simplify further and use compound angle formula for cosine in order to reduce its power to one.

**Formula used:**

Sine and cosine formula \[2{\sin ^2}x = 1 - \cos 2x\]

Algebraic identities for expansion of

${(a + b)^2} = {a^2} + 2ab + {b^2}$

${(a - b)^2} = {a^2} - 2ab + {b^2}$

Compound angle formula of cosine \[2{\cos ^2}x = 1 + \cos 2x\]

**Complete step by step answer:**

In order to rewrite the expression ${\sin ^8}x$ in terms of first power of cosine we will first convert sine into cosine as follows

${\sin ^8}x$

With the help of law of indices for brackets we can write it as

${\left( {{{\sin }^2}x} \right)^4}$

Dividing and multiplying it by $16$ we will get

$\dfrac{{16{{\left( {{{\sin }^2}x} \right)}^4}}}{{16}} \\

\Rightarrow\dfrac{1}{{16}}{\left( {2{{\sin }^2}x} \right)^4}\;\;\;\;\left[ {\because {2^4} = 16} \right] \\ $

We know that \[2{\sin ^2}x = 1 - \cos 2x\] , so replacing it with cosine and also again using the law of indices for brackets, we will get

\[\dfrac{1}{{16}}{\left( {{{\left( {1 - \cos 2x} \right)}^2}} \right)^2}\]

Using the algebraic identity ${(a - b)^2} = {a^2} - 2ab + {b^2}$ to open the parentheses

\[\dfrac{1}{{16}}{\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)^2}\]

Taking $(1 - 2\cos x)\;{\text{as}}\;a\;{\text{and}}\;{\cos ^2}2x\;{\text{as}}\;b$ and using the identity ${(a + b)^2} = {a^2} + 2ab + {b^2}$ to expand it more

\[\dfrac{1}{{16}}{\left( {\left( {1 - 2\cos 2x} \right) + {{\cos }^2}2x} \right)^2} \\

\dfrac{1}{{16}}\left( {{{\left( {1 - 2\cos 2x} \right)}^2} + 2\left( {1 - \cos 2x} \right){{\cos }^2}2x + {{\left( {{{\cos }^2}2x} \right)}^2}} \right) \\ \]

Simplifying this further with help of algebraic identities and compound cosine formula,

\[\dfrac{1}{{16}}\left( {1 - 2 \times 2\cos 2x + {{\left( {2\cos 2x} \right)}^2} + 2{{\cos }^2}2x - 4{{\cos }^3}x + {{\left( {\dfrac{{2{{\cos }^2}2x}}{2}} \right)}^2}} \right) \\

\Rightarrow\dfrac{1}{{16}}\left( {1 - 4\cos 2x + 4{{\cos }^2}2x + 2{{\cos }^2}2x - 4{{\cos }^3}x + {{\left( {\dfrac{{1 + \cos 4x}}{2}} \right)}^2}} \right)\;\;\;\;\left[ {\because 2{{\cos }^2}x = 1 + \cos 2x} \right] \\

\Rightarrow\dfrac{1}{{16}}\left( {1 - 4\cos 2x + 6{{\cos }^2}2x - \left( {3\cos 2x + \cos 6x} \right) + {{\left( {\dfrac{{1 + \cos 4x}}{2}} \right)}^2}} \right)\;\;\;\;\left[ {\because 4{{\cos }^3}x = 3\cos x + \cos 3x} \right]\; \\

\Rightarrow\dfrac{1}{{16}}\left( {1 - 4\cos 2x + 3\left( {1 + \cos 4x} \right) - \left( {3\cos 2x + \cos 6x} \right) + \left( {\dfrac{{1 + {{\cos }^2}4x + 2\cos 4x}}{4}} \right)} \right)\;\;\;\left[ {\because 2{{\cos }^2}x = 1 + \cos 2x} \right] \\

\Rightarrow\dfrac{1}{{16}}\left( {1 - 4\cos 2x + 3 + 3\cos 4x - 3\cos 2x - \cos 6x + \left( {\dfrac{{1 + {{\cos }^2}4x + 2\cos 4x}}{4}} \right)} \right)\; \\

\Rightarrow\dfrac{1}{{16}}\left( {4 - 7\cos 2x + 3\cos 4x - \cos 6x + \left( {\dfrac{{2\left( {1 + {{\cos }^2}4x + 2\cos 4x} \right)}}{{2 \times 4}}} \right)} \right)\; \\

\Rightarrow\dfrac{1}{{16}}\left( {4 - 7\cos 2x + 3\cos 4x - \cos 6x + \left( {\dfrac{{2 + 2{{\cos }^2}4x + 4\cos 4x}}{8}} \right)} \right) \\

\Rightarrow\dfrac{1}{{16}}\left( {4 - 7\cos 2x + 3\cos 4x - \cos 6x + \left( {\dfrac{{2 + (1 + \cos 8x) + 4\cos 4x}}{8}} \right)} \right)\;\;\;\;\left[ {\because 2{{\cos }^2}x = 1 + \cos 2x} \right] \\

\Rightarrow\dfrac{1}{{16}}\left( {\dfrac{{32 - 56\cos 2x + 24\cos 4x - 8\cos 6x + 3 + \cos 8x + 4\cos 4x}}{8}} \right) \\

\Rightarrow\dfrac{1}{{16 \times 8}}\left( {35 - 56\cos 2x + 28\cos 4x - 8\cos 6x + \cos 8x} \right) \\

\therefore\dfrac{1}{{128}}\left( {35 - 56\cos 2x + 28\cos 4x - 8\cos 6x + \cos 8x} \right) \\ \]

**Therefore, \[\dfrac{1}{{128}}\left( {35 - 56\cos 2x + 28\cos 4x - 8\cos 6x + \cos 8x} \right)\] is the simplified form of ${\sin ^8}x$ in terms of first power of cosine.**

**Note:**This is a lengthy and complex expansion so take care of the powers of the functions, signs (positive and negative), also where you are extra multiplying a factor so check there if you have divided with the same factor or not and write formulas in the line where it is being used.

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