Given that $cot x = \frac{2}{5}$ $cotx = 2/5$ , then what is the value of $tan (2 x)$ $tan(2x)$ , $cos (2 x)$ $cos(2x)$ and $sin (2 x)$ $sin(2x)$ ?

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Answer 1 · 2017-03-27T20:48:25

$tan 2 x = - \frac{20}{21}$ $tan2x =-20/21$
$sin 2x \ = 20/29$
$cos 2x \ = -21/29$

$cotx=2/5 => tanx=5/2$

Consider the following right angle triangle:

By elementary trigonometry we have:

$tan theta = "opp"/"adj" = 5/2 => theta = x$

By Pythagoras:

$\ \ \ \ h^2 = 2^2+5^2$
$:. h^2 = 4+25$
$:. h^2 = 29$
$:. \ \h = sqrt(29)$

And so;

$sin x = "opp"/"hyp" = 5/h = 5/sqrt(29)$
$cos x = "adj"/"hyp" = 2/h = 2/sqrt(29)$

Using the identity $tan(2theta) -= (2tan theta)/(1-tan^2theta)$ we have:

$tan2x = (2tan x)/(1-tan^2 x)$
$" " = (2*5/2)/(1-(5/2)^2)$
$" " = 5/(1-25/4)$
$" " = 5/(-21/4)$
$" " = -20/21$

And similarly using $sin2x=2sinxcosx$

$sin 2x = 2*5/sqrt(29)*2/sqrt(29)$
$" " = 20/29$

And similarly using $cos2x=cos^2x-sin^2x$

$cos 2x = (2/sqrt(29))^2 - (5/sqrt(29))^2$
$" " = 4/29 - 25/29$
$" " = -21/29$

Given that cotx = 2/5cotx=25cotx = 2/5, then what is the value of tan(2x)tan(2x)tan(2x), cos(2x)cos(2x)cos(2x) and sin(2x)sin(2x)sin(2x)?