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264.
Svođenje na oštar ugao

TEKST ZADATKA

Uprostiti izraz:

\frac {\sin^4{(\pi+x)}-\cos^4{(\pi-x)}} {\cos^2{\big(\frac {\pi} 2 +x\big)}-\sin^2{\big(x-\frac {\pi} 2 \big)}} - \frac {\sin^3{(\pi-x)}+\cos^3{(x-2\pi)}} {\cos{\big(x-\frac {\pi} 2 \big)}+\sin{\big(x+\frac {\pi} 2 \big)}}

REŠENJE ZADATKA

Svesti trigonometrijske funkcije na oštar ugao:

\frac {(-\sin{x})^4-(-\cos{x})^4} {(-\sin{x})^2-(-\cos{x})^2} - \frac {\sin^3{x}+\cos^3{x}} {\sin{x}+\cos{x}}

DODATNO OBJAŠNJENJE

\sin{(\pi+\alpha)}=-\sin{\alpha}

\cos{(\pi-\alpha)}=-\cos{\alpha}

\cos{\bigg(\frac {\pi} 2 +\alpha \bigg)}=-\sin{\alpha}

\sin{\bigg( \alpha - \frac {\pi} 2 \bigg)}=-\cos{\alpha}

\sin{(\pi-\alpha)}=\sin{\alpha}

\cos{(\alpha \pm 2k\pi)}=\cos{\alpha}, k \in Z

\cos{\bigg(\alpha-\frac {\pi} 2 \bigg)}=\sin{\alpha}

\sin{\bigg(\alpha+\frac {\pi} 2 \bigg)}=\cos{\alpha}

Srediti izraz.

\frac {\sin^4{x}-\cos^4{x}} {\sin^2{x}-\cos^2{x}} - \frac {\sin^3{x}+\cos^3{x}} {\sin{x}+\cos{x}}

Primeniti formulu za zbir kubova: $a^3+b^3=(a+b)(a^2-ab+b^2)$

\frac {\sin^4{x}-\cos^4{x}} {\sin^2{x}-\cos^2{x}} - \frac {(\sin{x}+\cos{x})(\sin^2x-\sin{x}\cos{x}+\cos^2x)} {\sin{x}+\cos{x}}

Primeniti formulu za razliku kvadrata: $a^2-b^2=(a-b)(a+b)$

\frac {(\sin^2{x}-\cos^2{x})(\sin^2x+\cos^2x)} {\sin^2{x}-\cos^2{x}} - \frac {(\sin{x}+\cos{x})(\sin^2x-\sin{x}\cos{x}+\cos^2x)} {\sin{x}+\cos{x}}

Primeniti osnovnu relaciju između trigonometrijskih funkcija: $\sin^2{\alpha}+\cos^2{\alpha}=1$

\frac {(\sin^2{x}-\cos^2{x})*1} {\sin^2{x}-\cos^2{x}} - \frac {(\sin{x}+\cos{x})(1-\sin{x}\cos{x})} {\sin{x}+\cos{x}}

Skratiti zajedničke činioce:

\frac {\cancel{(\sin^2{x}-\cos^2{x})}*1} {\cancel{\sin^2{x}-\cos^2{x}}} - \frac {\cancel{(\sin{x}+\cos{x})}(1-\sin{x}\cos{x})} {\cancel{\sin{x}+\cos{x}}}=1 - (1-\sin{x}\cos{x})=1 - 1+\sin{x}\cos{x}=\sin{x}\cos{x}

264.Svođenje na oštar ugao