Question

Oblicz [sin(alfa-pi)*cos(5pi/2 - alfa)]/[cos(3pi/2 +alfa)*sin(5pi-alfa)]

Answer 1

Wzory:

[tex]\sin\alpha=\sin\left(\alpha+2k\pi\right),\qquad k\in\mathbbb{Z}[/tex]

[tex]\cos\alpha=\cos\left(\alpha+2k\pi\right),\qquad k\in\mathbbb{Z}[/tex]

Wzory redukcyjne

[tex]\sin(\pi+\alpha)=-\sin\alpha[/tex]

[tex]\sin(\pi-\alpha)=\sin\alpha[/tex]

[tex]\cos\left(\dfrac{1}{2}\pi-\alpha\right)=\sin\alpha[/tex]

[tex]\cos\left(\dfrac{3}{2}\pi+\alpha\right)=\sin\alpha[/tex]

Rozwiązanie:

Założenie:

[tex]\alpha\neq k\pi,\qquad k\in\mathbbb{Z}[/tex]

[tex]\dfrac{\sin\left(\alpha-\pi\right)\cdot\cos\left(\dfrac{5\pi}{2}-\alpha\right)}{\cos\left(\dfrac{3\pi}{2}+\alpha\right)\cdot\sin\left(5\pi-\alpha\right)}=\dfrac{\sin\left(\alpha-\pi+2\pi\right)\cdot\cos\left(2\dfrac{1}{2}\pi-\alpha-2\pi\right)}{\cos\left(\dfrac{3}{2}\pi+\alpha\right)\cdot\sin\left(5\pi-\alpha-4\pi\right)}=[/tex]

[tex]=\dfrac{\sin\left(\alpha+\pi\right)\cdot\cos\left(\dfrac{1}{2}\pi-\alpha\right)}{\cos\left(\dfrac{3}{2}\pi+\alpha\right)\cdot\sin\left(\pi-\alpha\right)}=\dfrac{-\sin\alpha\cdot\sin\alpha}{\sin\alpha\cdot\sin\alpha}=\dfrac{-1\cdot1}{1\cdot1}=-1[/tex]