Simplifying the scalar product of $f(t)=\sum_{n=0}^7(c_n\cos\left(nt\right))$ with itself ($\langle f, f\rangle$)

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How can simplify this equation so that

$$\left\langle\sum _{n=0}^7c_n\cos\left(nt\right),\:\sum \:_{n=0}^7c_n\cos\left(nt\right)\:\right\rangle$$

becomes

$$\pi \left|c_0\right|^2+\frac{\pi \:}{2}\sum \:_{n=1}^7\left|c_n\right|^2$$

My attempt

$$\left\langle\sum _{n=0}^7c_n\cos\left(nt\right),\:\sum \:_{n=0}^7c_n\cos\left(nt\right)\:\right\rangle$$

$$=\sum \:_{n=0}^7\left(\sum \:\:_{n=0}^7\left|c_n\right|^2\langle \cos\left(nt\right),\:\cos\left(nt\right)\rangle\right)$$

$$=\sum \:_{n=0}^7\left(\sum \:\:_{n=0}^7\left|c_n\right|^2\int _{_0}^{\pi }\cos^2\left(nt\right)dt\:\right)$$

$$=\sum \:_{n=0}^7\left(\sum \:\:_{n=0}^7\left|c_n\right|^2\frac{\pi }{2}\right)$$

I don't know how to go on from here.