Wiedzac, ze cos 2x=m, oblicz: a) sin⁶x + cos⁶x b) sin⁸x + cos⁸x Prosze o wytlumaczenie i pelne rozwiazanie.

[latex]cos 2x=m[/latex] a) [latex]sin^6x + cos^6x=[/latex] [latex](sin^2x)^3 + (cos^2x)^3=[/latex] [latex](sin^2x+cos^2)(sin^4x-sin^2x cos^2x+cos^4x)=[/latex] [latex]sin^4x-sin^2x cos^2x+cos^4x=[/latex] [latex]sin^4x-2sin^2x cos^2x+cos^4x+sin^2x cos^2x=[/latex] [latex](cos^2x-sin^2x)^2+sin^2x cos^2x=[/latex] [latex](cos2x)^2+ frac{1}{4} cdot 4sin^2x cos^2x=[/latex] [latex]cos^22x^2+ frac{1}{4}cdot (2sinx cosx)^2=[/latex] [latex]cos^22x^2+ frac{1}{4}cdot (sin2x)^2=[/latex] [latex]cos^22x+ frac{1}{4}cdot sin^22x=[/latex] [latex]cos^22x+ frac{1}{4}cdot(1-cos^22x)=[/latex] [latex]m^2+ frac{1}{4}cdot(1-m^2)=[/latex] [latex]m^2+ frac{1}{4}-frac{1}{4}m^2=[/latex] [latex]frac{3}{4}m^2+ frac{1}{4}[/latex] ================ b) [latex]sin^8x + cos^8x=[/latex] [latex](sin^4x - cos^4x)^2+2sin^4x cos^4x=[/latex] [latex][sin^2x+cos^2x)(sin^2x-cos^2x)]^2+ frac{1}{8} cdot 16sin^4x cos^4x=[/latex] [latex](sin^2x-cos^2x)^2+ frac{1}{8} cdot (2sinx cosx)^4=[/latex] [latex](cos^2x-sin^2x)^2+ frac{1}{8} cdot (sin2x)^4=[/latex] [latex](cos2x)^2+ frac{1}{8} cdot (sin^22x)^2=[/latex] [latex]cos^22x+ frac{1}{8} cdot (1-cos^22x)^2=[/latex] [latex]m^2+ frac{1}{8} cdot (1-m^2)^2=[/latex] [latex]m^2+ frac{1}{8} cdot (1-2m^2+m^4)=[/latex] [latex]m^2+ frac{1}{8}- frac{1}{4} m^2+ frac{1}{8}m^4=[/latex] [latex]frac{1}{8}m^4+frac{3}{4} m^2+ frac{1}{8}[/latex]