Value of Trigonometric ratio (1 Viewer)

juantheron · Feb 6, 2016

$Calculation of $\displaystyle \frac{\tan 20^0+\tan 40^0+\tan 80^0-\tan 60^0}{\sin 40^0}$

$What i have try::, Using If $A+B+C= \pi\;,$ Then $\tan(A+B) = \tan(\pi-C)$

$So we get $\tan A+\tan B+\tan C = \tan A\cdot \tan B\cdot \tan C$

$And Using $\tan (60^0-A)\cdot \tan(A)\cdot \tan(60^0+A) = \tan (3A)$

$So our ratio is convert into $\displaystyle \frac{\tan 20^0+\underbrace{\tan 40^0+\tan 60^0+\tan 80^0}-2\tan 60^0}{\sin 40^0}$

$So $\displaystyle \frac{\tan 20^0-2\sqrt{3}+\tan 40^0\cdot \tan60^0\cdot \tan80^0}{\sin 40^0} = \frac{1}{\sin 40^0}\left[\tan 20^0-2\sqrt{3}+\frac{\sqrt{3}}{\tan 20^0}\right]$

Now How can I proceed further, Help me

Thanks

InteGrand · Feb 6, 2016

juantheron said:
$Calculation of $\displaystyle \frac{\tan 20^0+\tan 40^0+\tan 80^0-\tan 60^0}{\sin 40^0}$

$What i have try::, Using If $A+B+C= \pi\;,$ Then $\tan(A+B) = \tan(\pi-C)$

$So we get $\tan A+\tan B+\tan C = \tan A\cdot \tan B\cdot \tan C$

$And Using $\tan (60^0-A)\cdot \tan(A)\cdot \tan(60^0+A) = \tan (3A)$

$So our ratio is convert into $\displaystyle \frac{\tan 20^0+\underbrace{\tan 40^0+\tan 60^0+\tan 80^0}-2\tan 60^0}{\sin 40^0}$

$So $\displaystyle \frac{\tan 20^0-2\sqrt{3}+\tan 40^0\cdot \tan60^0\cdot \tan80^0}{\sin 40^0} = \frac{1}{\sin 40^0}\left[\tan 20^0-2\sqrt{3}+\frac{\sqrt{3}}{\tan 20^0}\right]$

Now How can I proceed further, Help me

Thanks

There might be a better way, but keep in mind that it is possible to tediously find the values of sin (and hence cos and then tan) of all integer angles from 1° to 90°. This is described for example at this page: http://www.intmath.com/blog/mathematics/how-do-you-find-exact-values-for-the-sine-of-all-angles-6212.

Here's the PDF on that page with exact values for sines: http://intmstat.com/blog/2011/06/exact-values-sin-degrees.pdf . You can use these for the angles in question. (The capital I's in that PDF's formulas refers to i, i.e. the "square root of -1".)

Paradoxica · Feb 6, 2016

Paradoxica · Feb 6, 2016

Value of Trigonometric ratio (1 Viewer)

juantheron

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InteGrand

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Paradoxica

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