Matematică, întrebare adresată de TyxW, 8 ani în urmă

Calculati sin 11pi/12 * cos 23pi/12.
* e inmultire​

Răspunsuri la întrebare

Răspuns de norishor90
8

sin(11pi/12)×cos(23pi/12)

1/2×[sin(17pi/6)+sin(-pi)]

1/2×(1/2+0)

1/2×1/2

(1×1)/(2×2)

1/4

Răspuns de adrianalitcanu2018
17

Răspuns:

 \frac{1}{4}

Explicație pas cu pas:

Calculam \sin{\frac{11\pi}{12}} :

\sin{\frac{11\pi}{12}}=\sin{\frac{12\pi-\pi}{12}}=\sin({\pi-\frac{\pi}{12}})=\sin\pi*\cos\frac{\pi}{12}-\cos\pi*\sin\frac{\pi}{12}=-\cos\pi*\sin\frac{\pi}{12}=-*(-1)*\sin\frac{\pi}{12}=\sin\frac{\pi}{12}

Vedem cat face \sin\frac{\pi}{12}:

\sin\frac{\pi}{12}=\sin({\frac{\pi}{3}-\frac{\pi}{4}}})=\sin{\frac{\pi}{3}}*\cos{\frac{\pi}{4}}-\sin{\frac{\pi}{4}}*\cos{\frac{\pi}{3}}=\frac{\sqrt3}{2}*\frac{\sqrt2}{2}-\frac{\sqrt2}{2}*\frac{1}{2}=\frac{\sqrt6-\sqrt2}{4}

Procedam la fel cu  \cos{\frac{23\pi}{12}} :

\cos{\frac{23\pi}{12}}=\cos{\frac{24\pi-\pi}{12}}=\cos({\frac{24\pi}{12}-\frac{\pi}{12}})=\cos(2\pi-\frac{\pi}{12})=\cos2\pi*\cos{\frac{\pi}{12}}+\sin2\pi*\sin{\frac{\pi}{12}}=\cos{\frac{\pi}{12}}

Vedem cat face \cos\frac{\pi}{12}:

\cos\frac{\pi}{12}=\cos({\frac{\pi}{3}-\frac{\pi}{4}}})=\cos{\frac{\pi}{3}}*\cos{\frac{\pi}{4}}+\sin{\frac{\pi}{4}}*\sin{\frac{\pi}{3}}=\frac{1}{2}*\frac{\sqrt2}{2}+\frac{\sqrt2}{2}*\frac{\sqrt3}{2}=\frac{\sqrt6+\sqrt2}{4}

Inmultim acum cele doua rezultate si avem:

\sin\frac{\pi}{12}*\cos\frac{\pi}{12}=\frac{\sqrt6-\sqrt2}{4}*\frac{\sqrt6+\sqrt2}{4}=\frac{6-2}{16}=\frac{4}{16}=\frac{1}{4}

Alte întrebări interesante