Find all x and y between 0 and 2pi so that the equation is ture
-2 cosx -i sin y = 2 sinx +i
i am also having trouble with this one
Simplify the power of i
i to the 39th power
i^1 = i
i^2 = -1
i^3 = -i
i^4 = (i^2)^2 =-1^2 = 1
i^38 =(i^2)^19 = -1
so
i^39 = -1*i = -i
If -2 cosx -i sin y = 2 sinx +i
both the real and imaginary terms on opposite sides of the equation must be equal.
-2 cos x = 2 sin x
Divide both sides by 2 cos x to solve for x
-1 = tan x
x = 3 pi/4 and 7 pi/4
From the imaginary part of the equation
sin y = -1
y = (3/2) pi
To find all values of x and y between 0 and 2pi that satisfy the equation -2cos(x) - i sin(y) = 2sin(x) + i, we can approach it in the following steps:
Step 1: Separate the real and imaginary parts of the equation.
-2cos(x) = 2sin(x)
and
-sin(y) = 1
Step 2: Simplify each equation separately.
For the first equation, divide both sides by 2:
cos(x) = -sin(x)
For the second equation, multiply both sides by -1:
sin(y) = -1
Step 3: Solve each equation to find the values of x and y within the given range.
For the first equation, we use the identity cos(x) = sin(pi/2 - x):
sin(pi/2 - x) = -sin(x)
To find solutions for x, we compare the angles inside the trigonometric functions:
pi/2 - x = -x + k * pi, where k is an integer
Simplifying further:
pi/2 = -k * pi
x = pi/2 - k * pi
Since the range for x is between 0 and 2pi, we have two solutions:
x = pi/2 and x = 3pi/2
For the second equation, we have y = 3pi/2 as the only solution.
Therefore, the solutions for x and y between 0 and 2pi that satisfy the given equation are x = pi/2, x = 3pi/2, and y = 3pi/2.
Regarding the second question, simplifying i to the power of 39 can be done by considering the pattern in powers of i.
i raised to any positive power follows a pattern: i, -1, -i, 1.
Since 39 is not divisible by 4, we can divide it by 4 and find the remainder:
39 divided by 4 is 9, with a remainder of 3.
Since 3 corresponds to the third term in the sequence, the solution will be -i.
Therefore, i to the power of 39 simplifies to -i.