Considere que A=sen 240º e que B=cos 150º . O valor do produto A.B vale:( resposta com cálculo)
a) ( )3/4
b) ( ) - √3/4
c) ( ) - 1/4
d) ( )1/2
e) ( )√3/2
Soluções para a tarefa
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Olá!
![\\ \mathsf{A \cdot B =} \\\\ \mathsf{\sin 240^o \cdot \cos 150^o} = \\\\ \mathsf{\sin (180^o + 60^o) \cdot \cos (90^o + 60^o)} = \\\\ \mathsf{ \left (\sin 180^o \cdot \cos 60^o + \sin 60^o \cdot \cos 180^o \right ) \cdot \left (\cos 90^o \cdot \cos 60^o - \sin 90^o \cdot \sin 60^o \right )} = \\\\ \left [ 0 \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot (- 1)\right ] \cdot \left ( 0 \cdot \frac{1}{2} - 1 \cdot \frac{\sqrt{3}}{2} \right ) = \\\\ \mathsf{- \frac{\sqrt{3}}{2} \cdot - \frac{\sqrt{3}}{2}}} = \\\\ \boxed{\mathsf{\frac{3}{4}}}](https://tex.z-dn.net/?f=%5C%5C+%5Cmathsf%7BA+%5Ccdot+B+%3D%7D+%5C%5C%5C%5C+%5Cmathsf%7B%5Csin+240%5Eo+%5Ccdot+%5Ccos+150%5Eo%7D+%3D+%5C%5C%5C%5C+%5Cmathsf%7B%5Csin+%28180%5Eo+%2B+60%5Eo%29+%5Ccdot+%5Ccos+%2890%5Eo+%2B+60%5Eo%29%7D+%3D+%5C%5C%5C%5C+%5Cmathsf%7B+%5Cleft+%28%5Csin+180%5Eo+%5Ccdot+%5Ccos+60%5Eo+%2B+%5Csin+60%5Eo+%5Ccdot+%5Ccos+180%5Eo+%5Cright+%29+%5Ccdot+%5Cleft+%28%5Ccos+90%5Eo+%5Ccdot+%5Ccos+60%5Eo+-+%5Csin+90%5Eo+%5Ccdot+%5Csin+60%5Eo+%5Cright+%29%7D+%3D+%5C%5C%5C%5C+%5Cleft+%5B+0+%5Ccdot+%5Cfrac%7B1%7D%7B2%7D+%2B+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D+%5Ccdot+%28-+1%29%5Cright+%5D+%5Ccdot+%5Cleft+%28+0+%5Ccdot+%5Cfrac%7B1%7D%7B2%7D+-+1+%5Ccdot+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D+%5Cright+%29+%3D+%5C%5C%5C%5C+%5Cmathsf%7B-+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D+%5Ccdot+-+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%7D%7D+%3D+%5C%5C%5C%5C+%5Cboxed%7B%5Cmathsf%7B%5Cfrac%7B3%7D%7B4%7D%7D%7D "\\ \mathsf{A \cdot B =} \\\\ \mathsf{\sin 240^o \cdot \cos 150^o} = \\\\ \mathsf{\sin (180^o + 60^o) \cdot \cos (90^o + 60^o)} = \\\\ \mathsf{ \left (\sin 180^o \cdot \cos 60^o + \sin 60^o \cdot \cos 180^o \right ) \cdot \left (\cos 90^o \cdot \cos 60^o - \sin 90^o \cdot \sin 60^o \right )} = \\\\ \left [ 0 \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot (- 1)\right ] \cdot \left ( 0 \cdot \frac{1}{2} - 1 \cdot \frac{\sqrt{3}}{2} \right ) = \\\\ \mathsf{- \frac{\sqrt{3}}{2} \cdot - \frac{\sqrt{3}}{2}}} = \\\\ \boxed{\mathsf{\frac{3}{4}}}")
LuanyPerys:
mt obg
Por nada!!
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