Answer:
a. cosθ = ¹/₂[e^jθ + e^(-jθ)] b. sinθ = ¹/₂[e^jθ - e^(-jθ)]
Step-by-step explanation:
a.We know that
e^jθ = cosθ + jsinθ and
e^(-jθ) = cosθ - jsinθ
Adding both equations, we have
e^jθ = cosθ + jsinθ
+
e^(-jθ) = cosθ - jsinθ
e^jθ + e^(-jθ) = cosθ + cosθ + jsinθ - jsinθ
Simplifying, we have
e^jθ + e^(-jθ) = 2cosθ
dividing through by 2 we have
cosθ = ¹/₂[e^jθ + e^(-jθ)]
b. We know that
e^jθ = cosθ + jsinθ and
e^(-jθ) = cosθ - jsinθ
Subtracting both equations, we have
e^jθ = cosθ + jsinθ
-
e^(-jθ) = cosθ - jsinθ
e^jθ + e^(-jθ) = cosθ - cosθ + jsinθ - (-jsinθ)
Simplifying, we have
e^jθ - e^(-jθ) = 2jsinθ
dividing through by 2 we have
sinθ = ¹/₂[e^jθ - e^(-jθ)]
Answer:
a=18
Step-by-step explanation:
Answer:
Yes it has infinity of solutions.
Step-by-step explanation:
If you plug in certain numbers it will make the statements true.
Answer:
X=7
(x - 3 )(x +5)
(x+3) (x-5)
X - 35
Step-by-step explanation:
(x + 1)2 = 16
Expanding , we have
2x + 2 = 16
2x = 16 - 2
2x = 14
X= 7.
Remark
Let the pigs = p
Let the chickens = c
Givens
30 heads is another way of saying that there were 30 creatures in the barnyard. In addition each chicken has 2 legs and each pig has 4
c + p = 30
2C + 4P = 100 (2) Divide (2) by 2
C + 2P = 50 (3) Subtract (1) from (3)
<u>C + P = 30</u>
P = 20
So there were 20 pigs.
