←Vrati se na zadatke3. ZadatakAko je sinx+cosx=a\sin x+\cos x=asinx+cosx=a, ∣a∣≤2|a|\le\sqrt2∣a∣≤2, koliko je 1+cos2xctgx2−tgx2 ?\frac{1+\cos2x}{\operatorname{ctg}\frac{x}{2}-\operatorname{tg}\frac{x}{2}}\ ?ctg2x−tg2x1+cos2x ?Prikaži rješenjeRješenjeZapišimo izraz 1+cos2xctgx2−tgx2\frac{1+\cos2x}{\operatorname{ctg}\frac{x}{2}-\operatorname{tg}\frac{x}{2}}ctg2x−tg2x1+cos2x u jednostavnijem obliku: 1+cos2xctgx2−tgx2=1+cos2x1+cosxsinx−1−cosxsinx=(1+2cos2x−1)sinx1+cosx−1+cosx=2cos2xsinx2cosx=cosxsinx.\begin{aligned} \frac{1+\cos2x}{\operatorname{ctg}\frac{x}{2}-\operatorname{tg}\frac{x}{2}} &=\frac{1+\cos2x}{\frac{1+\cos x}{\sin x}-\frac{1-\cos x}{\sin x}}\\ &=\frac{(1+2\cos^2x-1)\sin x}{1+\cos x-1+\cos x}\\ &=\frac{2\cos^2x\sin x}{2\cos x}=\cos x\sin x. \end{aligned}ctg2x−tg2x1+cos2x=sinx1+cosx−sinx1−cosx1+cos2x=1+cosx−1+cosx(1+2cos2x−1)sinx=2cosx2cos2xsinx=cosxsinx. Slijedi sinx+cosx=a⟹(sinx+cosx)2=a2\sin x+\cos x=a \quad\Longrightarrow\quad (\sin x+\cos x)^2=a^2sinx+cosx=a⟹(sinx+cosx)2=a2 ⟹1+2sinxcosx=a2\Longrightarrow\quad 1+2\sin x\cos x=a^2⟹1+2sinxcosx=a2 ⟹sinxcosx=a2−12.\Longrightarrow\quad \sin x\cos x=\frac{a^2-1}{2}.⟹sinxcosx=2a2−1.Pomakni formulu lijevo ili desno.
