Think De Moivre

Geometry Level 2

Let a a and b b be positive integers such that 1 cos ⁡ ( 8 0 ∘ ) + i sin ⁡ ( 8 0 ∘ ) \dfrac1{\cos(80^\circ) + i \sin(80^\circ) } can be written in the form of cos ⁡ ( a ) − i sin ⁡ ( b ) \cos(a) - i \sin(b) , where a a and b b are minimized, calculate a − b a - b .

Note : Angles are measured in degrees.


The answer is 0.

Hamza A by Hamza A , 19, USA

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4 solutions

Michael Fuller
Feb 13, 2016

1 cos ⁡ ( 80 ° ) + i sin ⁡ ( 80 ° ) = ( 1 cos ⁡ ( 80 ° ) + i sin ⁡ ( 80 ° ) ) ( cos ⁡ ( 80 ° ) − i sin ⁡ ( 80 ° ) cos ⁡ ( 80 ° ) − i sin ⁡ ( 80 ° ) ) = cos ⁡ ( 80 ° ) − i sin ⁡ ( 80 ° ) cos ⁡ 2 ( 80 ° ) + sin ⁡ 2 ( 80 ° ) = cos ⁡ ( 80 ° ) − i sin ⁡ ( 80 ° ) \cfrac { 1 }{ \cos { \left( 80° \right) } +i\sin { \left( 80° \right) } } \\ =\left( \cfrac { 1 }{ \cos { \left( 80° \right) } +i\sin { \left( 80° \right) } } \right) \left( \cfrac { \cos { \left( 80° \right) } -i\sin { \left( 80° \right) } }{ \cos { \left( 80° \right) } -i\sin { \left( 80° \right) } } \right) \\ =\cfrac { \cos { \left( 80° \right) } -i\sin { \left( 80° \right) } }{ \cos ^{ 2 }{ \left( 80° \right) } +\sin ^{ 2 }{ \left( 80° \right) } } \\ =\cos { \left( 80° \right) } -i\sin { \left( 80° \right) }

Therefore a = b = 80 ° ⇒ a − b = 0 a=b=80° \Rightarrow a-b= \large \color{#20A900}{\boxed{0}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Mateus Gomes
Feb 12, 2016

cos ⁡ ( 4 π 9 ) + i sin ⁡ ( 4 π 9 ) = e ( i 4 π 9 ) \cos(\frac{4\pi}{9})+i\sin(\frac{4\pi}{9})=\large{e}^{(\frac{i4\pi}{9})} 1 cos ⁡ ( 4 π 9 ) + i sin ⁡ ( 4 π 9 ) = [ e ( i 4 π 9 ) ] − 1 = e − ( i 4 π 9 ) = cos ⁡ ( 4 π 9 ) − i s i n ( 4 π 9 ) \frac{1}{\cos(\frac{4\pi}{9})+i\sin(\frac{4\pi}{9})}=\Large[{e}^{(\frac{i4\pi}{9})}]^{-1}=\Large{e}^{-(\frac{i4\pi}{9})}=\cos(\frac{4\pi}{9})-isin(\frac{4\pi}{9}) A − B = 0 A-B=\color{#3D99F6}{\boxed{\Large 0}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Somyaneel Sinha
Feb 14, 2016

To define a complex no. with modulus 1 a must be equal to b.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

why must a=b?

Pi Han Goh - 5 years, 3 months ago

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surely that must be its argument and the answer must appear as cosx +isinx thus a-b = x-x =0

Somyaneel Sinha - 5 years, 3 months ago
. .
Feb 25, 2021

1 cos ⁡ ( 80 ° ) + i sin ⁡ ( 80 ° ) = 1 × cos ⁡ ( 80 ° ) + i sin ⁡ ( 80 ° ) → cos ⁡ ( a ) − i sin ⁡ ( b ) → a = 80 , b = 80 → a − b = 80 − 80 = 0 . \displaystyle \frac { 1 } { \cos ( 80 \degree ) + i \sin ( 80 \degree ) } = 1 \times \cos ( 80 \degree ) + i \sin ( 80 \degree ) \rightarrow \cos ( a ) - i \sin ( b ) \rightarrow a = 80, b = 80 \rightarrow a - b = 80 - 80 = \boxed { 0 } .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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