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

1 Answer

Answer:

# tan2x =-20/21 #
# sin 2x \ = 20/29 #
# cos 2x \ = -21/29 #

Explanation:

# cotx=2/5 => tanx=5/2 #

Consider the following right angle triangle:

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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 #