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Please help me with my Trigonometry?
Write each expression in the standard form for a complex number, a + bi.
1. [2(cos(π/20))+isin(π/20)]^5
2. [√1(cos(3π/14))+isin(3π/14)]^7
3. [√5(cos(5π/16))+isin(5π/16)]^4
4. [^3√7(cos(π/18))+isin(π/18)]^6
5. (1 + i)^5
6. (-√3+i)5
1 Answer
- 2 years ago
(R * (cos(t) + i * sin(t)))^n = R^n * (cos(n * t) + i * sin(n * t)
(2 * (cos(pi/20) + i * sin(pi/20)))^5 =>
2^5 * (cos(5pi/20) + i * sin(5pi/20)) =>
32 * (cos(pi/4) + i * sin(pi/4)) =>
32 * (sqrt(2)/2) * (1 + i) =>
16 * sqrt(2) * (1 + i) =>
16 * sqrt(2) + 16 * sqrt(2) * i
You can handle the next 3
(-sqrt(3) + i)^5 =>
(-sqrt(3) + i * 1)^5
R * cos(t) = -sqrt(3)
R * sin(t) = 1
R^2 * cos(t)^2 + R^2 * sin(t)^2 = (-sqrt(3))^2 + 1^2
R^2 * (cos(t)^2 + sin(t)^2) = 3 + 1
R^2 * 1 = 4
R^2 = 4
R = -2 , 2
Let's use R = 2
(-sqrt(3) + i)^5 =>
(2 * (-sqrt(3)/2 + i * (1/2))^5 =>
2^5 * (-sqrt(3)/2 + i * (1/2))^5
-sqrt(3)/2 = cos(t)
t = arccos(-sqrt(3)/2)
t = 5pi/6 , 7pi/6
sin(t) = 1/2
t = arcsin(1/2)
t = pi/6 , 5pi/6
The common angle is 5pi/6
2^5 * (-sqrt(3)/2 + i * (1/2))^5 =>
32 * (cos(5pi/6) + i * sin(5pi/6))^5 =>
32 * (cos(25pi/6) + i * sin(25pi/6)) =>
32 * (cos(2pi + pi/6) + i * sin(2pi + pi/6)) =>
32 * (cos(pi/6) + i * sin(pi/6)) =>
32 * (sqrt(3)/2 + i * 1/2) =>
16 * sqrt(3) + 16 * i
Now you can handle #5
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