MHB What is the solution for d in a trigonometric equation with multiple terms?

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The discussion focuses on finding the value of d in the equation $\sin^7 x = a\sin 7x + b\sin 5x + c\sin 3x + d\sin x$. Participants are encouraged to refer to the Problem of the Week guidelines for submission procedures. Two members, lfdahl and castor28, provided correct solutions to the problem. The thread highlights the collaborative nature of solving complex trigonometric equations. Engaging with such mathematical challenges fosters a deeper understanding of trigonometric identities and their applications.
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Here is this week's POTW:

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Find $d$ when $\sin^7 x=a\sin 7x+b\sin 5x+c\sin 3x+d\sin x$.

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Congratulations to the following members for their correct solution!(Cool)

1. lfdahl
2. castor28

Solution from lfdahl:
One way to solve the problem is to use the orthogonality of sine functions:

If $n$ and $m$ are positive integers, then

\[\int_{0}^{2\pi}\sin mx \sin nx dx = \frac{1}{2}\int_{0}^{2\pi}\left ( \cos (m-n)x - \cos (m+n)x\right )dx =\left\{\begin{matrix} 0, \;\;\;m \ne n\\ \pi, \;\;\; m = n \end{matrix}\right.\]Thus multiplying by $\sin x$ and integrating over the interval $[0;2\pi]$, we get

\[\sin^7x = a\sin 7x + b \sin 5x + c \sin 3x + d \sin x \]

\[\Rightarrow \int_{0}^{2\pi}\sin^8x dx = a\int_{0}^{2\pi}\sin 7x \sin x dx +b\int_{0}^{2\pi}\sin 5x \sin x dx +a\int_{0}^{2\pi}\sin 3x \sin x dx +d\int_{0}^{2\pi}\sin^2x dx \\\\ =d\int_{0}^{2\pi}\sin^2x dx \\\\ = \pi \cdot d\]

It remains to validate the definite integral of $\sin^8x$. In doing so, we can use the power reduction formula:

\[\int_{0}^{2\pi}\sin^nx dx = \frac{n-1}{n}\int_{0}^{2\pi}\sin^{n-2}xdx - \left [ \frac{1}{n}\cos x \sin^{n-1}x \right ]^{2\pi}_0 = \frac{n-1}{n}\int_{0}^{2\pi}\sin^{n-2}xdx\]

- which can be easily shown by integration by parts. Thus, we end up with:

\[d= \frac{1}{\pi}\int_{0}^{2\pi}\sin^8x dx = \frac{1}{\pi}\frac{7}{8}\int_{0}^{2\pi}\sin^6x dx = \frac{1}{\pi}\frac{7\cdot 5}{8\cdot6}\int_{0}^{2\pi}\sin^4x dx = \frac{1}{\pi}\frac{7\cdot 5 \cdot 3}{8\cdot6 \cdot 4}\int_{0}^{2\pi}\sin^2x dx \\\\ = \frac{7\cdot 5 \cdot 3}{8\cdot 6 \cdot 4} = \frac{35}{64}.\]

Alternate solution from castor28:
We write the identity for a few values of $x$ (in degrees):
$$
\begin{array}{ll}
x=30:\qquad&-\dfrac{a}{2}+\dfrac{b}{2}+c+\dfrac{d}{2}=\dfrac{1}{128}\\
x=45: &\dfrac{\sqrt{2}}{2}(-a-b+c+d)=\left(\dfrac{\sqrt{2}}{2}\right)^7=\dfrac{\sqrt{2}}{16}\\
x=60:&\dfrac{\sqrt{3}}{2}(a-b+d)=\left(\dfrac{\sqrt{3}}{2}\right)^7 = \dfrac{27\sqrt{3}}{128}\\
x=90:&-a+b-c+d=1
\end{array}
$$
After simplification, we get the system of linear equations:
\begin{align*}
-a+b+2c+d&=\dfrac{1}{64}\\
-a-b+c+d&=\dfrac18\\
a-b+d&=\dfrac{27}{64}\\
-a+b-c+d&=1
\end{align*}

whose solution is $a=-\dfrac{1}{64}$, $b=\dfrac{7}{64}$, $c=-\dfrac{21}{64}$, $\bf d=\dfrac{35}{64}$.
 
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