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1d) If y = $2^x$ $(sin)^2$ xcosx find $y_n$.
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| written 7.5 years ago by |
$2^x$ = $e^{xlog2}$ = $e^{ax}$ where a = log2 $\frac{2(sin)^2 xcosx)}{2}$ = $\frac{(sin)^1 xcosx.sinx ×2)}{2}$ = $\frac{(sinx.sin2x)}{2}$ = $\frac{(2sinx.sin2x)}{(2×2)}$ = $\frac{cosx}{4}$ - $\frac{cos3x}{4}$ ∴$sin^2$ xcosx = $\frac{cosx}{4}$ - $\frac{cos3x}{4}$ Y = $\frac{(e^{ax} cosx)}{4}$ - $\frac{(e^{ax} cos3x)}{4}$ $y_n$ = $\frac{1}{4}(r_1)^n e^{ax}$ cos(x+n$φ_1$) - $\frac{1}{4}(r_2)^n e^{ax}$ cos(3x+n$φ_2$) $y_n$ = $\frac{1}{4}(r_1)^n 2^{1x}$ cos(x+n$φ_1$) - $\frac{1}{4}(r_2)^n 2^{1x}$ cos(3x+n$φ_2$) $r_1$ = $((log2)^2)^{\frac{1}{2}}$ + 1$r_2$ = $((log2)^2)^{\frac{1}{2}}$ +$3^2$ $φ_1$ = $tan^{-1}$ [$\frac{1}{log2}$] $φ_2$ = $tan^{-1}$[$\frac{3}{log2}$]
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