relativity1
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I)prove cosθ-isinθ=cis(-θ)
II)Prove sinθ-icosθ=cis(θ-pi/2)
II)Prove sinθ-icosθ=cis(θ-pi/2)
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Argand proofs (1 Viewer)
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relativity1
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Wait can i just say for I)since its in the 4th quadrant angle is just -theta?
Drsoccerball
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Thats basically what it is ... and for ii its just basic trig switch.
godofindolence
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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)
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godofindolence
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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.