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Question: Expand \({\sin ^6}x\)?...

Expand sin⁡6x{\sin ^6}x?

Explanation

Solution

In this question first we split sin⁡6x{\sin ^6}x into multiples of sin⁡2x{\sin ^2}x terms. Then we have to use the formula of 2sin⁡2x=1−cos⁡2x2{\sin ^2}x = 1 - \cos 2x to break the question into higher angles of cos⁡x\cos x. We have to apply different formulas till we do not get the degree one of trigonometric function in the result.

Complete answer:
In the above question, sin⁡6x{\sin ^6}x is a given function.
=sin⁡6x= {\sin ^6}x
We can also write it as
=(sin⁡2x)(sin⁡2x)(sin⁡2x)= \left( {{{\sin }^2}x} \right)\left( {{{\sin }^2}x} \right)\left( {{{\sin }^2}x} \right)
Now we will use the formula 2sin⁡2x=1−cos⁡2x2{\sin ^2}x = 1 - \cos 2x
=(1−cos⁡2x2)(1−cos⁡2x2)(1−cos⁡2x2)= \left( {\dfrac{{1 - \cos 2x}}{2}} \right)\left( {\dfrac{{1 - \cos 2x}}{2}} \right)\left( {\dfrac{{1 - \cos 2x}}{2}} \right)
Now multiplying the constants in denominator and first two brackets
=18(1−cos⁡2x−cos⁡2x+cos⁡22x)(1−cos⁡2x)= \dfrac{1}{8}\left( {1 - \cos 2x - \cos 2x + {{\cos }^2}2x} \right)\left( {1 - \cos 2x} \right)
=18(1−2cos⁡2x+cos⁡22x)(1−cos⁡2x)= \dfrac{1}{8}\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)\left( {1 - \cos 2x} \right)
Now, multiplying both the brackets
=18(1−cos⁡2x−2cos⁡2x+2cos⁡22x+cos⁡2x−cos⁡32x)= \dfrac{1}{8}\left( {1 - \cos 2x - 2\cos 2x + 2{{\cos }^2}2x + {{\cos }^2}x - {{\cos }^3}2x} \right)
=18(1−3cos⁡2x+3cos⁡22x−cos⁡32x)= \dfrac{1}{8}\left( {1 - 3\cos 2x + 3{{\cos }^2}2x - {{\cos }^3}2x} \right)
Now we will use the identity cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4{\cos ^3}x - 3\cos x
In the above identity, in place of x put 2x and bring cos⁡3x{\cos ^3}x one side.
We get cos⁡32x=cos⁡6x+3cos⁡2x4{\cos ^3}2x = \dfrac{{\cos 6x + 3\cos 2x}}{4}
=18(1−3cos⁡2x+3cos⁡22x−cos⁡6x4−3cos⁡2x4)= \dfrac{1}{8}\left( {1 - 3\cos 2x + 3{{\cos }^2}2x - \dfrac{{\cos 6x}}{4} - \dfrac{{3\cos 2x}}{4}} \right)
Now on taking LCM, we get
=18(4−12cos⁡2x+12cos⁡22x−cos⁡6x−3cos⁡2x4)= \dfrac{1}{8}\left( {\dfrac{{4 - 12\cos 2x + 12{{\cos }^2}2x - \cos 6x - 3\cos 2x}}{4}} \right)
On simplification, we get
=132(4−15cos⁡2x+12cos⁡22x−cos⁡6x)= \dfrac{1}{{32}}\left( {4 - 15\cos 2x + 12{{\cos }^2}2x - \cos 6x} \right)
Now using the formula 2cos⁡2x=1+cos⁡2x2{\cos ^2}x = 1 + \cos 2x
=132(4−15cos⁡2x+6(2cos⁡22x)−cos⁡6x)= \dfrac{1}{{32}}\left( {4 - 15\cos 2x + 6\left( {2{{\cos }^2}2x} \right) - \cos 6x} \right)
We have to use angle 2x instead of x in the formula.
=132(4−15cos⁡2x+6(1+cos⁡4x)−cos⁡6x)= \dfrac{1}{{32}}\left( {4 - 15\cos 2x + 6\left( {1 + \cos 4x} \right) - \cos 6x} \right)
=132(4−15cos⁡2x+6+6cos⁡4x−cos⁡6x)= \dfrac{1}{{32}}\left( {4 - 15\cos 2x + 6 + 6\cos 4x - \cos 6x} \right)
Finally, on simplification we get
=132(10−15cos⁡2x+6cos⁡4x−cos⁡6x)= \dfrac{1}{{32}}\left( {10 - 15\cos 2x + 6\cos 4x - \cos 6x} \right)

Note: There can be many ways to do this question like we can use the complex numbers (euler’s formula) and can convert the sine and cosine functions into exponential functions. The third way is using De Moivre’s theorem in which we can express 2cos⁡(nx)=zn+1zn  and  2sin⁡(nx)=zn−1zn  2\cos \left( {nx} \right) = {z^n} + \dfrac{1}{{{z^n}}}\,\,and\,\,2\sin \left( {nx} \right) = {z^n} - \dfrac{1}{{{z^n}}}\,\,.