What Is the Laplace Transform of a Trigonometric Function?

The Laplace transform turns a time-domain function into a function of the complex variable $s$. For the standard sine and cosine functions, the results are rational expressions. The angular frequency matters: for example, the formulas for $\sin(4t)$ and $\cos(4t)$ use $\omega=4$, not the coefficient of $t$ squared in some unrelated expression.

The standard formulas

Using the one-sided Laplace transform,

$$\mathcal{L}\{f(t)\}(s)=\int_0^\infty e^{-st}f(t)\,dt,$$

the basic trig transforms are

$$\mathcal{L}\{\sin(\omega t)\}=\frac{\omega}{s^2+\omega^2}, \qquad \mathcal{L}\{\cos(\omega t)\}=\frac{s}{s^2+\omega^2}.$$

Here $\omega$ is a real angular frequency. For nonzero $\omega$, these integrals converge when $\operatorname{Re}(s)>0$. At $\omega=0$, $\sin(0t)$ is identically zero and its transform is $0$ for every $s$, while $\cos(0t)=1$ has transform $1/s$ for $\operatorname{Re}(s)>0$. If the frequency is written as $2\pi f$, use $\omega=2\pi f$ in the formulas.

A useful distinction is that sine has $\omega$ in the numerator, while cosine has $s$. Both have the same denominator, $s^2+\omega^2$.

Why the formulas hold

For real $s>0$, direct integration gives

$$\begin{aligned} \int_0^\infty e^{-st}\sin(\omega t)\,dt &=\left[\frac{e^{-st}(-s\sin(\omega t)-\omega\cos(\omega t))}{s^2+\omega^2}\right]_0^\infty\\ &=\frac{\omega}{s^2+\omega^2}, \end{aligned}$$

because the exponential term tends to zero as $t\to\infty$. Similarly,

$$\begin{aligned} \int_0^\infty e^{-st}\cos(\omega t)\,dt &=\left[\frac{e^{-st}(-s\cos(\omega t)+\omega\sin(\omega t))}{s^2+\omega^2}\right]_0^\infty\\ &=\frac{s}{s^2+\omega^2}. \end{aligned}$$

These formulas extend to complex $s$ in the regions where the integrals converge: $\operatorname{Re}(s)>0$ for nonzero $\omega$, and also for $\cos(0t)$. For $\sin(0t)=0$, the transform is $0$ for every $s$.

Worked example: an exponentially damped sine

Find the transform of $e^{-2t}\sin(4t)$. The exponential shift rule says

$$\mathcal{L}\{e^{at}f(t)\}(s)=F(s-a),$$

where $F(s)=\mathcal{L}\{f(t)\}(s)$. The region of convergence shifts accordingly: for a complex $a$, the condition $\operatorname{Re}(s)>0$ becomes $\operatorname{Re}(s)>\operatorname{Re}(a)$; for real $a$, this is $\operatorname{Re}(s)>a$. In this example, $a=-2$ and $f(t)=\sin(4t)$, so replace $s$ by $s+2$ in the sine formula:

$$\mathcal{L}\{e^{-2t}\sin(4t)\} =\frac{4}{(s+2)^2+16}.$$

The original integral converges when $\operatorname{Re}(s+2)>0$, or $\operatorname{Re}(s)>-2$. This shift in the convergence region is part of the result, not just an optional technical detail.

For comparison, the same rule gives

$$\mathcal{L}\{e^{3t}\cos(2t)\} =\frac{s-3}{(s-3)^2+4}, \qquad \operatorname{Re}(s)>3.$$

Other expressions involving trig functions

The basic formulas also help with some combinations. For example, use the identity $\sin^2(t)=(1-\cos(2t))/2$ and linearity:

$$\begin{aligned} \mathcal{L}\{\sin^2(t)\} &=\frac12\left(\frac{1}{s}-\frac{s}{s^2+4}\right)\\ &=\frac{2}{s(s^2+4)}, \qquad \operatorname{Re}(s)>0. \end{aligned}$$

Products such as $\sin(at)\cos(bt)$ can often be rewritten as sums using product-to-sum identities before taking the transform. For powers such as $t\sin(\omega t)$, a different useful rule is $\mathcal{L}\{t f(t)\}=-\frac{d}{ds}F(s)$, when the derivative and transform are valid.

Not every expression called a trigonometric function has a simple rational transform. For instance, $\tan(t)$ has poles on the positive time axis, so its ordinary integral from zero to infinity is not defined as a standard Laplace transform. Check that the function is integrable with the exponential weight before applying a table formula.

Common mistakes

• Using frequency instead of angular frequency: If the function is $\sin(2\pi f t)$, substitute $\omega=2\pi f$, not $f$.
• Dropping the frequency in the sine numerator: The numerator is $\omega$, so $\mathcal{L}\{\sin(4t)\}=4/(s^2+16)$.
• Shifting in the wrong direction: A factor $e^{at}$ changes $F(s)$ to $F(s-a)$. Thus damping by $e^{-ct}$, with $c>0$, produces $F(s+c)$.
• Ignoring the region of convergence: State the condition on $\operatorname{Re}(s)$ along with the expression, especially after an exponential shift.