PERIODIC signals is to get the harmonics. i.e, nω0 represents the nth harmonic of ω0Procedure to find the period of x1(t) + x2(t) + --------Step1: calculate the individual periods T1, T2, T3, T4 ⋯ ⋯ etcStep2: calculate the ratio like \(\frac{{{T_1}}}{{{T_2}}},\frac{{{T_1}}}{{{T_3}}},\frac{{{T_1}}}{{{T_4}}} \cdots \;etc\)Step3: if the ratios in step2 are rational then periodic.Step4: calculate LCM of denominators in step2Step5: T = LCM × T1Alternative methodCalculate the GCD or HCF of ( T1, T2, T3, T4 ⋯ ⋯ )\({T_1} = \frac{{2\pi }}{{{\omega _1}}},\;{T_2} = \frac{{2\pi }}{{{\omega _2}}} \cdots etc\)ω0 = GCD = 2π / T“T” is the REQUIRED period.\(sinAsinB = \frac{1}{2}\left[ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right]\)\(COSACOSB = \frac{1}{2}\left[ {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right]\)Calculation:Given SIGNAL is X(t) = cos (2 πt) cos (4 πt) – sin (2πt) sin (4πt)It is a combination of two signals.x1(t) = cos (2 πt) cos (4 πt)= ½ [cos(2πt – 4πt) + cos(2πt + 4πt)]= ½ [cos2πt + cos6πt]= ½ cos2πt + ½ cos6πtX2(t) = sin (2πt) sin (4πt)= ½[cos(2πt – 4πt) - cos(2πt + 4πt)]= ½ [cos2πt – cos6πt]= ½ cos2πt – ½ cos6πtNow x(t) will bex(t) = ½ cos2πt + ½ cos6πt –[ ½ cos2πt – ½ cos6πt]x(t) = cos6πtcomparing with x(t) = cosω0tω0 = 6πT = 2π / 6πT = 1/3 sec