What is the polar form of 1+sina+icosa
To find the polar form of 1 + sin(a) + i * cos(a), we need to convert it from rectangular form to polar form. The rectangular form of a complex number is a + bi, where a is the real part and b is the imaginary part.
To convert from rectangular to polar form, we use the following formula:
r = sqrt(a^2 + b^2)
theta = arctan(b/a)
For the given complex number, a = 1, b = sin(a), and b = cos(a), so we can substitute these values into the formula:
r = sqrt(1^2 + sin^2(a) + cos^2(a))
= sqrt(1 + sin^2(a) + cos^2(a))
= sqrt(2)
theta = arctan((sin(a))/1)
= arctan(sin(a))
Therefore, the polar form of 1 + sin(a) + i * cos(a) is sqrt(2) * (cos(arctan(sin(a))) + i * sin(arctan(sin(a)))).
To determine the polar form of the expression 1 + sin(a) + i*cos(a), we need to represent it in terms of a magnitude (r) and an angle (θ).
1. Express 1 + sin(a) + i*cos(a) as a complex number:
It can be written as z = 1 + sin(a) + i*cos(a).
2. Rewrite the expression in terms of Euler's formula:
Euler's formula states that e^(ix) = cos(x) + i*sin(x).
Let's rewrite the expression using Euler's formula to get:
z = 1 + e^(ia).
3. Convert the expression to polar form:
The polar form of a complex number is given by:
z = r*(cos(θ) + i*sin(θ)).
By comparing the expression obtained in step 2 with the polar form, we can say that:
r = 1 (magnitude)
θ = a (angle)
Therefore, the polar form of 1 + sin(a) + i*cos(a) is 1*(cos(a) + i*sin(a)), where r = 1 and θ = a.
Yyh
sina = cos(π/2-a)
cosa = sin(π/2-a)
1+sina+icosa
= 1+cos(π/2-a) + i sin(π/2-a)
= 1 + cis(π/2-a)
There is no simple cis(x) form because of that pesky 1.