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 Please find sin theta and tan theta if cos theta = -12/13 and theta lies in the third quadrant.

Indranil  Oct 20, 2017
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Using

 

sin^2 ( theta) + cos^2 (theta)  = 1    we have

 

sin^2(theta) +  ( -12 / 13)^2  = 1

 

sin^(theta)  =  1 - (-12/13)^2

 

sin^2(theta)  =  1 - 144/169

 

sin^2(theta)  =    169/169  -  144/169

 

sin^2 (theta)  =  25/169       take the square root of both sides

 

sin (theta)  = ± √ [25/169]     and the sine is negative in the 3rd quadrant

 

sin (theta)  =  -5/13

 

tan (theta)  =   sin (theta) / cos (theta)  =  ( -5/13) / (-12/13)  =  5/12

 

 

cool cool cool

CPhill  Oct 20, 2017

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