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Algebra Level 5

If the expression

( cis ( 3 ) ) cos ⁡ − 1 ( 3 5 ) (\text{cis}(3))^{\cos^{-1}\left(\frac{3}{5}\right)}

Can be represented as − a b + c b i \dfrac{-a}{b}+\dfrac{c}{b}i for positive coprime integers a , b , c a,b,c , find a + b + c a+b+c .

Details and assumptions :

It's up to you to decide whether the measurements are in degrees or radians :P

cis ( θ ) = cos ⁡ ( θ ) + i sin ⁡ ( θ ) \text{cis}(\theta)=\cos(\theta)+i\sin(\theta)


The answer is 286.

Trevor Arashiro by Trevor Arashiro , 22, USA

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1 solution

Trevor Arashiro
Mar 13, 2016

cis ( θ ) n = cis ( n θ ) \text{cis}(\theta)^n=\text{cis}(n\theta)

( cis ( 3 ∘ ) ) cos ⁡ − 1 ( 3 5 ) = cos ⁡ ( 3 ∘ cos ⁡ − 1 ( 3 5 ) ) + i sin ⁡ ( 3 ∘ cos ⁡ − 1 ( 3 5 ) ) (\text{cis}(3^{\circ}))^{\cos^{-1}\left(\frac{3}{5}\right)}=\cos\left(3^{\circ}\cos^{-1}\left(\frac{3}{5}\right)\right)+i\sin\left(3^{\circ}\cos^{-1}\left(\frac{3}{5}\right)\right)

Triple angle formula (think of 3 as a number and not an angle).

4 cos ⁡ 3 ( cos ⁡ − 1 ( 3 5 ) ) − 3 cos ⁡ ( cos ⁡ − 1 ( 3 5 ) ) + 3 i sin ⁡ ( cos ⁡ − 1 ( 3 5 ) ) − 4 sin ⁡ 3 ( cos ⁡ − 1 ( 3 5 ) ) 4\cos^3\left(\cos^{-1}\left(\frac{3}{5}\right)\right)-3\cos\left(\cos^{-1}\left(\frac{3}{5}\right)\right)+3i\sin\left(\cos^{-1}\left(\frac{3}{5}\right)\right)-4\sin^3\left(\cos^{-1}\left(\frac{3}{5}\right)\right)

cos ⁡ − 1 ( x ) = sin ⁡ − 1 ( 1 − x 2 ) \cos^{-1}(x)=\sin^{-1}\left(\sqrt{1-x^2}\right)

cos ⁡ − 1 ( 3 5 ) = sin ⁡ − 1 ( 4 5 ) \cos^{-1}\left(\frac{3}{5}\right)=\sin^{-1}\left(\frac{4}{5}\right)

4 ⋅ 27 125 − 3 ⋅ 3 5 + 3 i ⋅ 4 5 − 4 i ⋅ 64 125 4\cdot \frac{27}{125}-3\cdot \frac{3}{5}+3i\cdot \frac{4}{5}-4i\cdot \frac{64}{125}

− 117 125 + 44 125 i \boxed{\dfrac{-117}{125}+\dfrac{44}{125}i}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

It should be 3 3 radians, not 3 ∘ 3^\circ . The answer would be completely different if it's in degrees instead of radians.

Prasun Biswas - 5 years, 3 months ago

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As long as one is used through the entire problem, the answer comes out the same. It has to be for both 3 and acos.

Trevor Arashiro - 5 years, 3 months ago

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That doesn't seem to be the case. Here are the Wolfram|Alpha verification links: 1 and 2

Prasun Biswas - 5 years, 3 months ago

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@Prasun Biswas – That's cuz u have a c o s ( 3 / 5 ∘ ) acos(3/5^{\circ})

Try this

Wolfram refuses to associate inverse trig functions with degrees, so u have to convert rad to deg

Trevor Arashiro - 5 years, 3 months ago

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@Trevor Arashiro – Ah, so you meant to make the inverse cosine return the value in degrees instead of radians! Nice. But it's pretty unconventional. Nevertheless, as long as it works. +1

Prasun Biswas - 5 years, 3 months ago

i added 125 two times lol but i got it :D

Mardokay Mosazghi - 5 years, 2 months ago

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