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November 10th, 2016, 03:13 AM   #1
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Arctan identity

Hey I'm trying to answer this without a calculator.
Arctan(2+sqrt(3)) with calculator it is 75.

Last edited by skipjack; November 10th, 2016 at 11:09 PM.
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November 10th, 2016, 03:31 AM   #2
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I assume that is in degrees. You can reference a table of the unit circle to see that is the case. If you don't have a table that precise, use the half-angle formula of tangent for 150 degrees.

Last edited by Compendium; November 10th, 2016 at 03:33 AM.
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November 10th, 2016, 04:03 AM   #3
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tan(75°) = tan(45° + 30°) = (tan(45°) + tan(30°))/(1 - tan(45°)tan(30°)).

As tan(45°) = 1 and tan(30°)= 1/√3, tan(75°) = (√3 + 1)/(√3 - 1) = 2 + √3.
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November 10th, 2016, 04:28 AM   #4
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$0 < t < 90$

$\tan{t}=2+\sqrt{3} \implies \sin{t}=2\cos{t}+\sqrt{3}\cos{t}$

$2\cos{t}=\sin{t}-\sqrt{3}\cos{t}$

$\cos{t}=\dfrac{1}{2}\sin{t}-\dfrac{\sqrt{3}}{2}\cos{t}$

$\cos{t}=\sin(150)\sin{t}+\cos(150)\cos{t}$

$\cos{t}=\cos(150-t) \implies t=150-t \implies t = 75$
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November 10th, 2016, 10:09 PM   #5
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arctan.png
arctan(2 + √3) = 60° + 15° = 75°.
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November 11th, 2016, 07:49 PM   #6
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$$2+\sqrt3=\frac{\sin60+\sin90}{\sin150}= \frac{2\sin^275}{2\cos75\sin75}=\tan75\implies \arctan(2+ \sqrt3)=75^\circ$$
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Last edited by greg1313; November 11th, 2016 at 07:53 PM.
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November 11th, 2016, 11:05 PM   #7
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Using cot(A) = csc(2A) + cot(2A), 2 + √3 = csc(30°) + cot(30°) = cot(15°) = tan(75°).
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November 12th, 2016, 05:26 AM   #8
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