If y=sin2ax+cosax,prove that yn=a^n{1+(-1) sin2ax}^1/2
y=sinax+cosaxyn=ansin{nπ2+ax}+bncos{nπ2+ax}=an(sin{nπ2+ax}+cos{nπ2+ax})Since(sinax+cosax)=(1+sin2ax)12,thenyn=an{1+sin2(nπ2+ax)}=an{1+sin(2nπ2+2ax)}12={1+(−1)nsin2ax}12\begin{aligned} y &=\sin a x+\cos a x \\ y_{n} &=a^{n} \sin \left\{\frac{n \pi}{2}+a x\right\}+b^{n} \cos \left\{\frac{n \pi}{2}+a x\right\} \\ &=a^{n}\left(\sin \left\{\frac{n \pi}{2}+a x\right\}+\cos \left\{\frac{n \pi}{2}+a x\right\}\right) \end{aligned}\\ Since (\sin a x+\cos a x)=(1+\sin 2 a x)^{\frac{1}{2}} , then\\ \begin{aligned} y_{n} &=a^{n}\left\{1+\sin 2\left(\frac{n \pi}{2}+a x\right)\right\} \\ &=a^{n}\left\{1+\sin \left(\frac{2 n \pi}{2}+2 a x\right)\right\}^{\frac{1}{2}} \\ &=\left\{1+(-1)^{n} \sin 2 a x\right\}^{\frac{1}{2}} \end{aligned}yyn=sinax+cosax=ansin{2nπ+ax}+bncos{2nπ+ax}=an(sin{2nπ+ax}+cos{2nπ+ax})Since(sinax+cosax)=(1+sin2ax)21,thenyn=an{1+sin2(2nπ+ax)}=an{1+sin(22nπ+2ax)}21={1+(−1)nsin2ax}21
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