(M) Show thatis a linearly independent set of functions defined on. Use the method of Exercise 37. (This result will be needed in Exercise 34 in Section 4.5.)

Assume \({c_1} \cdot 1 + {c_2} \cdot \cos t + {c_3} \cdot {\cos ^2}t + ... + {c_7} \cdot {\cos ^6}t = 0\).

For \(t = 0,.1,.2,...,.6\), the above equation gives the system as shown below:

\(\left( {\begin{array}{*{20}{c}}1&{\cos 0}&{{{\cos }^2}0}&{{{\cos }^3}0}&{{{\cos }^4}0}&{{{\cos }^5}0}&{{{\cos }^6}0}\\1&{\cos .1}&{{{\cos }^2}.1}&{{{\cos }^3}.1}&{{{\cos }^4}.1}&{{{\cos }^5}.1}&{{{\cos }^6}.1}\\1&{\cos .2}&{{{\cos }^2}.2}&{{{\cos }^3}.2}&{{{\cos }^4}.2}&{{{\cos }^5}.2}&{{{\cos }^6}.2}\\1&{\cos .3}&{{{\cos }^2}.3}&{{{\cos }^3}.3}&{{{\cos }^4}.3}&{{{\cos }^5}.3}&{{{\cos }^6}.3}\\1&{\cos .4}&{{{\cos }^2}.4}&{{{\cos }^3}.4}&{{{\cos }^4}.4}&{{{\cos }^5}.4}&{{{\cos }^6}.4}\\1&{\cos .5}&{{{\cos }^2}.5}&{{{\cos }^3}.5}&{{{\cos }^4}.5}&{{{\cos }^5}.5}&{{{\cos }^6}.5}\\1&{\cos .6}&{{{\cos }^2}.6}&{{{\cos }^3}.6}&{{{\cos }^4}.6}&{{{\cos }^5}.6}&{{{\cos }^6}.6}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\\{{c_4}}\\{{c_5}}\\{{c_6}}\\{{c_7}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\\0\\0\\0\end{array}} \right)\)

That is, \(Ac = 0\).

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}1&{\cos 0}&{{{\cos }^2}0}&{{{\cos }^3}0}&{{{\cos }^4}0}&{{{\cos }^5}0}&{{{\cos }^6}0}\\1&{\cos .1}&{{{\cos }^2}.1}&{{{\cos }^3}.1}&{{{\cos }^4}.1}&{{{\cos }^5}.1}&{{{\cos }^6}.1}\\1&{\cos .2}&{{{\cos }^2}.2}&{{{\cos }^3}.2}&{{{\cos }^4}.2}&{{{\cos }^5}.2}&{{{\cos }^6}.2}\\1&{\cos .3}&{{{\cos }^2}.3}&{{{\cos }^3}.3}&{{{\cos }^4}.3}&{{{\cos }^5}.3}&{{{\cos }^6}.3}\\1&{\cos .4}&{{{\cos }^2}.4}&{{{\cos }^3}.4}&{{{\cos }^4}.4}&{{{\cos }^5}.4}&{{{\cos }^6}.4}\\1&{\cos .5}&{{{\cos }^2}.5}&{{{\cos }^3}.5}&{{{\cos }^4}.5}&{{{\cos }^5}.5}&{{{\cos }^6}.5}\\1&{\cos .6}&{{{\cos }^2}.6}&{{{\cos }^3}.6}&{{{\cos }^4}.6}&{{{\cos }^5}.6}&{{{\cos }^6}.6}\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}1&1&1&1&1&1&1\\1&{\cos .1}&{{{\cos }^2}.1}&{{{\cos }^3}.1}&{{{\cos }^4}.1}&{{{\cos }^5}.1}&{{{\cos }^6}.1}\\1&{\cos .2}&{{{\cos }^2}.2}&{{{\cos }^3}.2}&{{{\cos }^4}.2}&{{{\cos }^5}.2}&{{{\cos }^6}.2}\\1&{\cos .3}&{{{\cos }^2}.3}&{{{\cos }^3}.3}&{{{\cos }^4}.3}&{{{\cos }^5}.3}&{{{\cos }^6}.3}\\1&{\cos .4}&{{{\cos }^2}.4}&{{{\cos }^3}.4}&{{{\cos }^4}.4}&{{{\cos }^5}.4}&{{{\cos }^6}.4}\\1&{\cos .5}&{{{\cos }^2}.5}&{{{\cos }^3}.5}&{{{\cos }^4}.5}&{{{\cos }^5}.5}&{{{\cos }^6}.5}\\1&{\cos .6}&{{{\cos }^2}.6}&{{{\cos }^3}.6}&{{{\cos }^4}.6}&{{{\cos }^5}.6}&{{{\cos }^6}.6}\end{array}} \right|\\\det A \ne 0\end{array}\)

By the invertible matrix theorem, has only a trivial solution. Thus, is a linearly independent set of functions defined on .