Chapter 5.4, Problem 77E

To determine

To prove:The equation $y_1+y_2=2A\cos \frac{2\pi t}{T}\cos \frac{2\pi x}{\lambda }$ .

Explanation of Solution

Given information:

The given equations are,

  $y_1=A\cos 2\pi \left(\frac{t}{T}-\frac{x}{\lambda }\right)$

  $y_2=A\cos 2\pi \left(\frac{t}{T}+\frac{x}{\lambda }\right)$

Here, the amplitudeis $A$ , the time period is $T$ and the wave length is $\lambda$ .

Formula Used:

The sum formula is,

  $\cos \left(x+y\right)=\cos x\cos y-\sin x\sin y$

  $\cos \left(x-y\right)=\cos x\cos y+\sin x\sin y$

Proof:

Add equation $y_1$ and $y_2$ .

  $\begin{array}{c}{y}_1+y_2=A\cos 2\pi \left(\frac{t}{T}-\frac{x}{\lambda }\right)+A\cos 2\pi \left(\frac{t}{T}+\frac{x}{\lambda }\right)\\ =A\left[\cos \left(\frac{2\pi t}{T}-\frac{2\pi x}{\lambda }\right)+\cos \left(\frac{2\pi t}{T}+\frac{2\pi x}{\lambda }\right)\right]\end{array}$

Apply the sum and difference formula for the sine function.

  $\begin{array}{c}{y}_1+y_2=A\left[\cos \left(\frac{2\pi t}{T}-\frac{2\pi x}{\lambda }\right)+\cos \left(\frac{2\pi t}{T}+\frac{2\pi x}{\lambda }\right)\right]\\ =A\left[\begin{array}{l}\cos \left(\frac{2\pi t}{T}\right)\cos \left(\frac{2\pi x}{\lambda }\right)+\sin \left(\frac{2\pi t}{T}\right)\sin \left(\frac{2\pi x}{\lambda }\right)+\cos \left(\frac{2\pi t}{T}\right)\cos \left(\frac{2\pi x}{\lambda }\right)\\ -\sin \left(\frac{2\pi t}{T}\right)\sin \left(\frac{2\pi x}{\lambda }\right)\end{array}\right]\\ =A\left(\cos \left(\frac{2\pi t}{T}\right)\cos \left(\frac{2\pi x}{\lambda }\right)\cos \left(\frac{2\pi t}{T}\right)\cos \left(\frac{2\pi x}{\lambda }\right)\right)\\ =A\left(2\cos \left(\frac{2\pi t}{T}\right)\cos \left(\frac{2\pi x}{\lambda }\right)\right)\end{array}$

From the above, it is clear thatone side of the equation can be transformed into other side of the equation.

Hence, proved.