Express 16sin^5x in the form of a sinx+b sin3x+c sin 5x
Answers
Answer:
Since we know that:
[math]\sin{(3x)}=3\sin{(x)}-4\sin^3{(x)}[/math]
[math]\sin{(5x)}=5\sin{(x)}-20\sin^3{(x)}+16\sin^5{(x)}[/math]
we can write:
[math]16\sin^5{(x)}=a\sin{(x)}+b[3\sin{(x)}-4\sin^3{(x)}]+c[5\sin{(x)}-20\sin^3{(x)}+16\sin^5{(x)}][/math]
From this equation, the coefficients for
[math]\sin{(x)}[/math] on the right-hand side must be zero:
[math]0=a+3b+5c[/math]
the coefficients for [math]\sin^3{(x)}[/math] on the right-hand side must be zero:
[math]0=-4b-20c[/math]
and the coefficients for [math]\sin^5{(x)}[/math] on the right-hand side must be [math]16[/math]:
[math]16=16c[/math]
Solving these three equations simultaneously, working up from the last equation, we find that:
[math]c=1[/math]
[math]b=-5[/math]
[math]a=10[/math]
Therefore,
[math]16\sin^5{x}=10\sin{(x)}-5\sin{(3x)}+\sin{(5x)}[/math]
Let [math]y = e^{ix}[/math] so that [math]\frac{1}{y} = e^{-ix}[/math]. Then for some [math]k[/math], we have the following:
[math]\displaystyle y^k + \frac{1}{y^k} = e^{ikx} + e^{-ikx} = 2 \cos {kx} \ldots (1)[/math]
[math]\displaystyle y^k - \frac{1}{y^k} = e^{ikx} - e^{-ikx} = 2i \sin {kx} \ldots (2)[/math]
To obtain the expansion for [math]\sin^5{x}[/math], we need to set [math]k = 1[/math] in (2) and then raise both sides to the power of 5:
[math]\displaystyle 32i^5\sin^5{x} = \left(y-\frac{1}{y}\right)^5[/math]
(expand by the Binomial Theorem)
[math]\displaystyle= y^5 - 5y^3 + 10y - \frac{10}{y} + \frac{5}{y^3} - \frac{1}{y^5}[/math]
[math]\displaystyle = \left(y^5-\frac{1}{y^5}\right) - 5\left(y^3-\frac{1}{y^3}\right) + 10\left(y-\frac{1}{y}\right)[/math]
By setting k = 5, 3, 1 in (2) and substituting from those results, we now have:
[math]\displaystyle 32i\sin^5{x}= 2i \sin{5x} - 5(2i\sin{3x}) + 10(2i\sin{x})[/math]
[math]\displaystyle \therefore 16\sin^5{x} = \sin{5x} - 5\sin{3x} + 10\sin{x}.[/math]