Q38E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Prove the sum of angles formula for the sine function by following these steps. Fix $\mathbf{x}$ .

 

$\left(a\right)$  Let  $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{sin}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{t}\mathbf{)}$. Show that $\mathbf{f}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$, the standard sum of angles formula for $\mathbf{sin}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{t}\mathbf{)}$ . $\mathbf{f}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{sinx}$ , and $\mathbf{f}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{cosx}$.

 

 $\left(b\right)$ Use the auxiliary equation technique to solve the initial value problem $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{sinx}$, and $\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{cosx}$ 

 

$\left(c\right)$ By uniqueness, the solution in part $\left(b\right)$  is the same as following these steps. Fix $\mathbf{x}$. $\mathbf{f}\left(\mathbf{t}\right)$ from part $\left(a\right)$. Write this equality; this should be the standard sum of angles formula for sin$\left(x+t\right)$.

Step-by-Step Solution

Answer

1. Differentiating with respect to t we find that $\mathrm{f}\text{'}\left(\mathrm{t}\right)=\cos (\mathrm{x}+\mathrm{t})$  and $\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)=-\sin (\mathrm{x}+\mathrm{t})$   then show that $\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{f}\left(\mathrm{t}\right)=0,\mathrm{f}\left(0\right)=\mathrm{sinx}$  and $\mathrm{f}\text{'}\left(0\right)=\mathrm{cosx}$ .
2. The solution to the given initial value problem is  $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{sinx}\mathrm{cost}+\mathrm{cosx}\mathrm{sint}$.
3. By uniqueness of the solution of the initial value problems, one has that  $\sin (\mathrm{x}+\mathrm{t})=\mathrm{sinx}\mathrm{cost}+\mathrm{cosx}\mathrm{sint}$.

Step 1: Differentiate the given value of f

To prove that  $\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{f}\left(\mathrm{t}\right)=0$ and  $\mathrm{f}\text{'}\left(0\right)=\mathrm{cosx}$.

 

First, one will find the required derivatives: -

$\begin{array}{l}\mathrm{f}\text{'}\left(\mathrm{t}\right)=\frac{\mathrm{d}}{\mathrm{dt}}\left(\sin \right(\mathrm{x}+\mathrm{t}\left)\right)\\ =\cos (\mathrm{x}+\mathrm{t})\times \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{x}+\mathrm{t})\\ =\cos (\mathrm{x}+\mathrm{t})\\ \mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{f}\text{'}\left(\mathrm{t}\right))=\frac{\mathrm{d}}{\mathrm{dt}}\left(\cos \right(\mathrm{x}+\mathrm{t}\left)\right)\\ =-\sin (\mathrm{x}+\mathrm{t})\times \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{x}+\mathrm{t})\\ =-\sin (\mathrm{x}+\mathrm{t}).\end{array}$

Now we can prove that:

$\begin{array}{l} \mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{f}\left(\mathrm{t}\right)=\sin (\mathrm{x}+\mathrm{t})+(-\sin (\mathrm{x}+\mathrm{t}\left)\right)\\ =0\end{array}$

$\mathrm{f}\left(0\right)=\sin (\mathrm{x}+0)=\mathrm{sinx}$

And

 

 $\begin{array}{c}\mathrm{f}\text{'}\left(0\right)=\cos (\mathrm{x}+0)\\ =\mathrm{cosx}\end{array}$

 

 

Step 2: Finding the general solution.

The auxiliary equation is ${\mathrm{r}}^2+1=0$  and its roots are ${\mathrm{r}}_{1,2}=\pm \mathrm{i}$.

The general solution to this problem has the form of $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}$, where  ${\mathrm{r}}_{1,2}=\mathrm{\alpha }\pm \mathrm{\beta i}$.

In this case $\mathrm{\alpha }=0$  and  $\mathrm{\beta }=1$, so the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}$ 

 

One will find the constants ${\mathrm{c}}_1$  and  ${\mathrm{c}}_2$ from the initial conditions.

$\begin{array}{l}\mathrm{y}\left(0\right)={\mathrm{c}}_1\cos 0+{\mathrm{c}}_2\sin 0={\mathrm{c}}_1=\mathrm{sinx}\\ \mathrm{y}\left(\mathrm{t}\right)=\mathrm{sinxcost}+{\mathrm{c}}_2\mathrm{sint};\\ \mathrm{y}\text{'}\left(\mathrm{t}\right)=\frac{d}{dt}(\mathrm{sinxcost}+{\mathrm{c}}_2\mathrm{sint})=-\mathrm{sinxsint}+{\mathrm{c}}_2\mathrm{cost};\\ \mathrm{y}\text{'}\left(0\right)=-\mathrm{sinxsin}0+{\mathrm{c}}_2\cos 0={\mathrm{c}}_2=\mathrm{cosx};\\ \mathrm{y}\left(\mathrm{t}\right)=\mathrm{sinxcost}+\mathrm{cosxsint}.\end{array}$

 

 

Step 3: Solve the initial value problem.

The formula for the sum of angles is given as:

 

$\sin (\mathrm{x}+\mathrm{t})=\mathrm{sinx}\mathrm{cost}+\mathrm{cosx}\mathrm{sint}$ 

 

One recognizes this as the formula of sine of the sum of two angles. This equality stands from the uniqueness of the solution to the initial value problem.