Question

Prove that for any value of $t$, the point $\left(x,y\right)=\left(\cos ht,\sin ht\right)$lies on the hyperbola$x^2-y^2=1$. Bonus question: In fact, these points will always lie on the right-hand side of the hyperbola; why? (This exercise involves hyperbolic functions.)

Short answer

Thus it is proved that$x^2-y^2=1$where$\left(x,y\right)=\left(\cos ht,\sin ht\right)$.

Step-by-step solution

Step 1. Given Information

We are given $\left(x,y\right)=\left(\cos ht,\sin ht\right)$and we need to prove that$x^2-y^2=1$.

Step 2. Proving the statement

We know, $\sin hx=\frac{e^x-e^{-x}}{2},\cos hx=\frac{e^x+e^{-x}}{2}$.

$$\begin{gathered} x^2-y^2 \\ =\cos h^{2}t-\sin h^{2}t \\ ={\left(\frac{e^x+e^{-x}}{2}\right)}^2-{\left(\frac{e^x-e^{-x}}{2}\right)}^2 \\ =\frac{e^{2x}+2e^x{e}^{-x}+e^{-2x}}{4}-\frac{e^{2x}-2e^x{e}^{-x}+e^{-2x}}{4} \\ =\frac{4e^x{e}^{-x}}{4} \\ =e^x{e}^{-x} \\ =e^x\times \frac{1}{e^x} \\ =1 \end{gathered}$$