Create a vector and calculate the hyperbolic cosine of each value.

X = [0 pi 2*pi 3*pi];
Y = cosh(X)

Y = 1×4
103 ×

    0.0010    0.0116    0.2677    6.1958

Plot the hyperbolic cosine function over the domain $$-5\le x\le 5.$$

x = -5:0.01:5; 
y = cosh(x);
plot(x,y)
grid on

The hyperbolic cosine satisfies the identity $\cosh \left(\mathit{x}\right)=\frac{{\mathit{e}}^{\mathit{x}}+{\mathit{e}}^{-\mathit{x}}}{2}$. In other words, $\cosh \left(\mathit{x}\right)$ is the average of ${\mathit{e}}^{\mathit{x}}$ and ${\mathit{e}}^{-\mathit{x}}$. Verify this by plotting the functions.

Create a vector of values between -3 and 3 with a step of 0.25. Calculate and plot the values of cosh(x), exp(x), and exp(-x). As expected, the curve for cosh(x) lies between the two exponential curves.

x = -3:0.25:3;
y1 = cosh(x);
y2 = exp(x);
y3 = exp(-x);
plot(x,y1,x,y2,x,y3)
grid on
legend('cosh(x)','exp(x)','exp(-x)','Location','bestoutside')