Find power of a sum of sinusoids We got this question to solve:

Calculate the power of the signal:
$$s(t) = 8\cos\left(20\pi t-\frac \pi4\right) + 4\sin(15\pi t)$$

Now, I thought of two approaches :

Use Parseval theorem, so first find Fourier transform, then calculate the required integral, but then we get a impulse function in it and I don't know how to integrate square of $\delta$.

Using the normal formula to calculate the power of a signal, and again in that, we get 3 terms, $\sin^2$ which is easy, then $\cos^2$ again simple, and then $\sin\cdot \cos$, which I know how to solve, but will be lengthy and I think there should be some another proper way to solve this question.


Any guidance would be appreciated.
HINT:
For real-valued periodic signal $x$ with period $T_0$ Parseval gives you
$$
P=\frac 1T_0\int_{-T_0/2}^{T_0/2} x^2(t)\mathrm{d}t = \sum_{n=-\infty}^{\infty}\lvert c_n\rvert^2
$$
Where $c_n$'s are Fourier coefficients. You can then use the properties in Equations $(1)$ and $(2)$.
$$
\mathcal F \big\{A\cos\left(2\pi f_0 t + \phi\right)\big\} = \frac A2\big[e^{j\phi}\delta\left(f - f_0\right) + e^{-j\phi}\delta\left(f + f_0\right)\big]\tag{1}
$$
$$
\mathcal F \big\{A\sin\left(2\pi f_0 t + \phi\right)\big\} = j\frac A2\big[e^{-j\phi}\delta\left(f + f_0\right)-e^{j\phi}\delta\left(f - f_0\right)\big]\tag{2}
$$
EDIT:
For Equation $(1)$ for instance, applying Parseval above you have the following:
\begin{align}
P&amp;= \sum_{n=-\infty}^{\infty}\lvert c_n\rvert^2\\
&amp;= \lvert c_{-1}\rvert^2 +  \lvert c_{1}\rvert^2\\
&amp; = \frac{A^2}{4}\cdot 1  + \frac{A^2}{4}\cdot 1\tag{$\lvert e^{\pm j\phi} \rvert = 1$} \\
\end{align}