Question

Calcule as expressões

Answer 1

Explicação passo-a-passo:

Basta saber que:

cos0=1 cosπ/2=0 cosπ=-1 cos3π/2=0 cos2π=1

sen0=0 senπ/2=1 senπ=0 sen3π/2=-1 sen2π=0

sen30°=1/2 sen45°=✓2/2 sen60°=✓3/2

cos30°=✓3/2 cos45°=✓2/2 cos60°=1/2

a) cos2π+3cosπ-(1/2)sen(π/2)

=1+3•(-1)-(1/2)•1

=1-3-(1/2)

=-2-(1/2)

=-4/2-1/2

=(-4-1)/2

=-5/2

b) 2sen(π/4)-4sen(π/2)+6cos(3π/2)

=2sen(π/4)-4•1+6•0

=2sen(π/4)-4

Agora para sen(π/4), note que π/4=180°/4=45°.

Logo,

=2•sen45°-4

=2•(✓2/2)-4

=✓2-4

c)sen(2π-(5π/6))

Para isso,

2π-(5π/6)=12π/6-5π/6=(12π-5π)/6=7π/6

Note que 7π/6=(7•180°)/6=210°

Temos que descobrir quanto vale o seno de 210°:

Note que 210°=30°+180°

sen(210°)=sen(30°+180°)=sen(30°)cos(180°)+sen(180°)cos(30°)

(pois sen(a+b)=senacosb+senbcosa

180°=π

=sen(30°)cos(180°)+sen(180°)cos(30°)

=sen(30°)cos(π)+sen(π)cos(30°)

=1/2•(-1)+0•(✓3/2)

=-1/2

d) cos(sen 2π)=cos0=1

e)cos(2π+π/3)

Basta abrir a soma do cosseno, isto é, cos(a+b)=cosacosb-senasenb.

cos(2π+π/3)=cos(2π)cos(π/3)-sen(2π)sen(π/3)

π/3=180°/3=60°

cos(2π)cos(π/3)-sen(2π)sen(π/3)

=cos(2π)cos(60°)-sen(2π)sen(60°)

=1•(1/2)-0•(✓3/2)

=1/2

f) cos2π+cosπ/3

π/3=180°/3=60°

cos2π+cosπ/3=cos2π+cos60°=1+1/2=2/2+1/2=(2+1)/2=3/2

g) sen120°+sen240°+sen360°

Note que:

360°=2π

240°=180°+60°=π+60°

120°=90°+30°=(π/2)+30°

Logo,

sen120°+sen240°+sen360°

=sen(π/2+30°)+sen(π+60°)+sen(2π)

=sen(π/2)cos(30°)+sen(30°)cos(π/2)+sen(π)cos(60°)+sen(60°)cos(π)+sen(2π)

=1•(✓3/2)+(1/2)•0+0•(1/2)+(✓3/2)•(-1)+0

=✓3/2-(✓3/2)

=0

h) 3cos300°-2cos225°+cos120°

Note que:

300°=270°+30°=3π/2+30°

225°=180°+45°=π+45°

120°=30°+90°=30°+π/2

3cos300°-2cos225°+cos120°

=3cos(3π/2+30°)-2cos(π+45°)+cos(30°+π/2)

=3[cos(3π/2)cos(30°)-sen(3π/2)sen(30°)]-2[cos(π)cos(45°)-sen(π)sen(45°)]+cos(30°)cos(π/2)-sen(30°)sen(π/2)]

=3[0•(✓3/2)-(-1)•(1/2)]-2[(-1)•(✓2/2)-0•(✓2/2)]+(✓3/2)•0-(1/2)•1

=3[1/2]-2[-✓2/2]-1/2

=3•1/2-2•(-✓2/2)-1/2

=3/2+✓2-1/2

=(3-1)/2+✓2

=2/2+✓2

=1+✓2