arg(1+sinA - i cosA)
Answers
Answer:
hope this helps you
Step-by-step explanation:
We have z=(1−sinα)+icosα
⇒∣z∣=
(1−sinα)
2
+cos
2
α
⇒∣z∣=
1+sin
2
α−2sinα+cos
2
α
⇒∣z∣=
2−2sinα
⇒∣z∣=
2(1−sinα)
⇒arg(z)=tan
−1
(
1−sinα
cosα
)
⇒cosα=
1+tan
2
2
α
1−tan
2
2
α
⇒1−sinα=1−
1+tan
2
2
α
2tan
2
α
⇒1−sinα=
1+tan
2
2
α
(1−tan
2
α
)
2
Therefore, ⇒
1−sinα
cosα
=
(1−tan
2
α
)
2
1−tan
2
2
α
=
1−tan
2
α
1+tan
2
α
⇒arg(z)=tan
−1
(
1−tan
2
α
1+tan
2
α
)
⇒arg(z)=tan
−1
(tan(
4
π
+
2
α
))
⇒arg(z)=
4
π
+
2
α
Answer:
answer is below my questioner.
Step-by-step explanation:
The argument can be computed as argz=arctanIzRz+nπ, so that should give you a start. The term nπ comes from the fact that arctan has a period of π, rather than 2π - so you also need to determine the n that gives the correct argument within a 2π period. Hint: think about what argz must be to place z in the correct quadrant... – Tomas Aschan Apr 24 '13 at 22:46