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Find sin2x, cos2x,and tan2x in the given information. csc x=9, tan x <0

Guest Apr 5, 2017
 #1
avatar+7336 
+3

csc x = 9

 

sin x = \(\frac{1}{9}\)

 

*edit* If the tangent is negative, and the sin is positive, then the cosine must be negative.

cos x = \( -\sqrt{1-(\frac{1}{9})^2}=-\frac{4\sqrt5}{9}\)

 

tan x = \( \frac{\sin x}{\cos x}=\frac{1}{9} \div -\frac{4\sqrt5}{9} = -\frac{\sqrt5}{20} \)

 

So..using the double-angle formulas:

 

sin (2x) = 2 sin x cos x = \(2\cdot\frac{1}{9}\cdot-\frac{4\sqrt5}{9} \mathbf{=-\frac{8\sqrt5}{81}} \)

 

cos (2x) = 1 - 2 sin2 x = \( 1-2\cdot(\frac{1}{9})^2\mathbf{=\frac{79}{81}} \)

 

tan (2x) = \( \frac{2\tan x}{1-\tan^2 x}=\frac{2\cdot{ \frac{-\sqrt5}{20}}}{1- (-\frac{\sqrt5}{20})^2}=\frac{-\frac{\sqrt5}{10}}{\frac{395}{400}}\mathbf{=-\frac{8\sqrt5}{79}}\)

hectictar  Apr 5, 2017
edited by hectictar  Apr 5, 2017
 #2
avatar+7336 
+2

Actually the tangent one is wrong.. I did it if tan x > 0 blush

 

....Okay I think I fixed it now

hectictar  Apr 5, 2017
edited by hectictar  Apr 5, 2017
 #3
avatar+91380 
+1

 

Yeah....you did fix it, hectictar.......check your answer for tan(2x)  by  comparing it with

sin(2x) / cos(2x)  .......

 

 

cool cool cool

CPhill  Apr 5, 2017
edited by CPhill  Apr 5, 2017

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