$y= sin rx + cos rx \; \; \; \; \; ( \theta = rx) \\ \text {Using the result of nth derivative of sin ax and cos ax we get,} \\ \therefore y_n = r^n \bigg [ sin \bigg(rx + {n\pi \over 2} \bigg)+ cos \bigg( rx +{n\pi \over 2} \bigg) \bigg] \\ \therefore y_n = r^n \bigg[ \bigg [ sin \bigg(rx + {n\pi \over 2} \bigg)+ cos \bigg( rx +{n\pi \over 2} \bigg) \bigg]^2 \bigg]^{1 \over 2} \\ \therefore y_n = r^n \bigg [ sin^2 \bigg(rx + {n\pi \over 2} \bigg)+ cos^2 \bigg( rx +{n\pi \over 2} \bigg) +2 sin \bigg(rx + {n\pi \over 2} \bigg) cos \bigg(rx + {n\pi \over 2} \bigg) \bigg]^{1 \over 2} \\ =r^n \bigg[ 1 + 2 \; sin \; 2\bigg(rx + {n\pi \over 2} \bigg) \bigg]^{\frac12} \\ = r^n [1 + sin (2 rx+ n\pi)]^{\frac12} \\ \therefore y_n= r^n [1 + (-1)^n sin2rx]^{\frac12} \; \; \; \; \; (since \; sin(n\pi+ \theta )=(-1)^n sin \theta ) \\ \therefore y_n= r^n \sqrt{1 + (-1)^n sin2 \theta} \\ Hence \; proved.$