+0  
 
0
108
3
avatar

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

Guest Apr 5, 2017
Sort: 

3+0 Answers

 #1
avatar+4172 
+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+4172 
+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+75344 
+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

15 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details