Wykaż że...

[MnMK]

\(\displaystyle{ 2(1+cosx)-sin ^{2}x=4cos ^{4} \frac{x}{2}}\)

[bayo84]

Ja to tak rozwiazalem:

\(\displaystyle{ 2(1+cosx) - sin^2x = 2+2cosx-sin^2x = 2+2cosx-1+cos^2x = 1-2cosx+cos^2x = (1-cosx)^2 = (1- \frac{1-tg^2 \frac{x}{2} }{1+tg^2 \frac{x}{2}})^2 = ( \frac{1+tg^2 \frac{x}{2}-1+tg^2 \frac{x}{2}}{1+tg^2 \frac{x}{2}})^2 = ( \frac{2tg^2 \frac{x}{2}}{1+tg^2 \frac{x}{2}} )^2 = \frac{4tg^4 \frac{x}{2}}{1+2tg^2 \frac{x}{2}+tg^4 \frac{x}{2}} = \frac{4 \frac{sin^4 \frac{x}{2}}{cos^4 \frac{x}{2}} }{1+2 \frac{sin^2 \frac{x}{2}}{cos^2 \frac{x}{2}}+ \frac{sin^4 \frac{x}{2}}{cos^4 \frac{x}{2}} } = \frac{4 \frac{sin^4 \frac{x}{2}}{cos^4 \frac{x}{2}} }{ \frac{cos^4 \frac{x}{2}+2sin^2 \frac{x}{2}cos^2 \frac{x}{2}+sin^4 \frac{x}{2}}{cos^4 \frac{x}{2}} } = \frac{4sin^4 \frac{x}{2}}{(cos^2 \frac{x}{2}+sin^2 \frac{x}{2})^2} = \frac{4sin^4 \frac{x}{2}}{(cos^2 \frac{x}{2}+1-cos^2 \frac{x}{2})^2} = 4sin^4 \frac{x}{2}}\)

Wszystko faknie tylko, ze wyszedl mi \(\displaystyle{ 4sin^4 \frac{x}{2}}\) ...

[piasek101]

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