Chapter 7.6, Problem 45AYU

To determine

Derive a formula for $\sin \left(5\theta \right)$ as a fifth degree polynomial in the variable $\cos \theta$

Answer to Problem 45AYU

  $16{\sin }^5\theta -20{\sin }^3\theta +5\sin \theta$

Explanation of Solution

Given information:

Derive a formula for $\sin \left(5\theta \right)$

Calculation:

We know that $5\theta =3\theta +2\theta$

Using the double angle formulas, $\cos \left(2\theta \right)=2{\cos }^2\theta -1$ and $\sin \left(2\theta \right)=2\sin \theta \cos \theta$

  $\sin 3\theta \cos 2\theta +\cos 3\theta \sin 2\theta =\sin 3\theta \left(2{\cos }^2\theta -1\right)+\cos 3\theta \left(2\sin \theta \cos \theta \right)$

Now find the expression of $\sin 3\theta$ and $\cos 3\theta$

We know that $3\theta =2\theta +\theta$

Rewrite the expression into this form $\sin \left(2\theta +\theta \right)$ and $\cos \left(2\theta +\theta \right)$

Apply the formula for sine function

  $\sin \left(2\theta +\theta \right)=\sin 2\theta \cos \theta +\cos 2\theta \sin \theta$

Substituting the double angle formulas,

  $\sin 2\theta \cos \theta +\cos 2\theta \sin \theta =2\sin \theta \cos \theta \cos \theta +\left({\cos }^2\theta -{\sin }^2\theta \right)\sin \theta$

  $\begin{array}{l}=2\sin \theta {\cos }^2\theta +{\cos }^2\theta \sin \theta -{\sin }^3\theta \\ =3\sin \theta {\cos }^2\theta -{\sin }^3\theta \end{array}$

Use the variation ${\cos }^2\theta =1-{\sin }^2\theta$ in the above equation.

  $\sin 3\theta =3\sin \theta \left(1-{\sin }^2\theta \right)-{\sin }^3\theta$

  $\begin{array}{l}=3\sin \theta -3{\sin }^3\theta -{\sin }^3\theta \\ =3\sin \theta -4{\sin }^3\theta \end{array}$

Similarly, apply the same formula, for cosine function $\cos \left(2\theta +\theta \right)=\cos 2\theta cos\theta -\sin 2\theta \sin \theta$

  $\cos 2\theta cos\theta -\sin 2\theta \sin \theta =\left(2{\cos }^2\theta -1\right)\cos \theta -2\sin \theta \cos \theta \sin \theta$

Take out common factor, $\cos \theta$

  $\left(2{\cos }^2\theta -1\right)\cos \theta -2\sin \theta \cos \theta \sin \theta =\cos \theta \left(2{\cos }^2\theta -1-2{\sin }^2\theta \right)$

Rewrite the expression

  $\cos \theta \left(2{\cos }^2\theta -1-2{\sin }^2\theta \right)=\cos \theta \left(2{\cos }^2\theta -1-2\left(1-{\cos }^2\theta \right)\right)$

  $=\cos \theta \left(2{\cos }^2\theta -1-2+2{\cos }^2\theta \right)$

Simplify the expression,

  $\cos \theta \left(2{\cos }^2\theta -1-2+2{\cos }^2\theta \right)=\cos \theta \left(4{\cos }^2\theta -3\right)$

  $=4{\cos }^3\theta -3\cos \theta$

Substitute the values of $\sin 3\theta$ and $\cos 3\theta$

  $\sin 3\theta \cos 2\theta +\cos 3\theta \sin 2\theta$

  $\begin{array}{l}=\sin 3\theta \left(2{\cos }^2\theta -1\right)+\cos 3\theta \left(2\sin \theta \cos \theta \right)\\ =\left(3\sin \theta -4{\sin }^3\theta \right)\left(2{\cos }^2\theta -1\right)+\left(4{\cos }^3\theta -3\cos \theta \right)\left(2\sin \theta \cos \theta \right)\\ =6\sin \theta {\cos }^2\theta -3\sin \theta -8{\sin }^3\theta {\cos }^2\theta +4{\sin }^3\theta +8{\cos }^4\theta \sin \theta -6\sin \theta {\cos }^2\theta \\ =-3\sin \theta -8{\sin }^3\theta {\cos }^2\theta +4{\sin }^3\theta +8{\cos }^4\theta \sin \theta \end{array}$

Factor the common terms in the expression

  $\sin 3\theta \cos 2\theta +\cos 3\theta \sin 2\theta =4{\sin }^3\left(1-2{\cos }^2\theta \right)+\sin \theta \left(8{\cos }^4\theta -3\right)$

Use the variation ${\cos }^2\theta =1-{\sin }^2\theta$ to express $\sin \left(5\theta \right)$ in terms of $\sin \theta$

  $\sin 5\theta =4{\sin }^3\theta \left(1-2\left(1-{\sin }^2\theta \right)\right)+\sin \theta \left(8{\left(1-{\sin }^2\theta \right)}^2-3\right)$

  $\begin{array}{l}=4{\sin }^3\theta \left(2{\sin }^2\theta -1\right)+\sin \theta \left(8+8{\sin }^4\theta -16{\sin }^2\theta -3\right)\\ =8{\sin }^5\theta -4{\sin }^3\theta +8\sin \theta +8{\sin }^5\theta -16{\sin }^3\theta -3\sin \theta \\ =16{\sin }^5\theta -20{\sin }^3\theta +5\sin \theta \end{array}$