relativity1
New Member
- Joined
- Feb 1, 2015
- Messages
- 23
- Gender
- Male
- HSC
- 2016
I)prove cosθ-isinθ=cis(-θ)
II)Prove sinθ-icosθ=cis(θ-pi/2)
II)Prove sinθ-icosθ=cis(θ-pi/2)
-
Looking for HSC notes and resources? Check out our Notes & Resources page
Argand proofs (1 Viewer)
- Thread starter relativity1
- Start date
relativity1
New Member
- Joined
- Feb 1, 2015
- Messages
- 23
- Gender
- Male
- HSC
- 2016
Wait can i just say for I)since its in the 4th quadrant angle is just -theta?
Drsoccerball
Well-Known Member
- Joined
- May 28, 2014
- Messages
- 3,647
- Gender
- Undisclosed
- HSC
- 2015
Thats basically what it is ... and for ii its just basic trig switch.
godofindolence
Member
- Joined
- Sep 17, 2015
- Messages
- 71
- Gender
- Male
- HSC
- 2015
1) Consider the values of sin and cos around the different quadrants of the unit circle (A-S-T-C) It's obvious that cosθ = cos-θ and sin-θ = sin-θ (for mod arg form, θ should be in the first 2 quadrants)
2) sin (pi/2-θ)= cos θ and similarly, cos (pi/2-θ) = sin θ (sum and difference of two angles)
2) sin (pi/2-θ)= cos θ and similarly, cos (pi/2-θ) = sin θ (sum and difference of two angles)
Last edited:
godofindolence
Member
- Joined
- Sep 17, 2015
- Messages
- 71
- Gender
- Male
- HSC
- 2015
haha oops, I've always thought that 3 and 4 quadrant would be -θ, note that $\mathrm{cis}(-\theta)=\cos (-\theta) + i\sin (-\theta) = \cos \theta - i\sin \theta$ (using even and odd properties of cosine and sine).$)
That would be the case if (and only if) θ is in quadrants 1 or 2, which it does not need to be.