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If cos(θ) = − 10/11 with θ in Quadrant III, what is sin(θ)?

Guest Oct 10, 2017

Best Answer 

 #1
avatar+7340 
+2

From the Pythagorean identity,

 

sin2 θ  +  cos2 θ     =  1        Plug in  -10/11  for  cos θ .

 

sin2 θ  +  (-10/11)2  =  1       \((-\frac{10}{11})^2=(-\frac{10}{11})(-\frac{10}{11})=\frac{100}{121}\)

 

sin2 θ  +  100/121   =  1       Subtract   100 / 121  from both sides.

 

sin2 θ   =   21/121                Since sin is negative in Quad III, take the negative sqrt of both sides.

 

sin θ     =  \(-\sqrt{\frac{21}{121}}\)

 

sin θ     =   \(-\frac{\sqrt{21}}{11}\)

hectictar  Oct 10, 2017
edited by hectictar  Oct 10, 2017
 #1
avatar+7340 
+2
Best Answer

From the Pythagorean identity,

 

sin2 θ  +  cos2 θ     =  1        Plug in  -10/11  for  cos θ .

 

sin2 θ  +  (-10/11)2  =  1       \((-\frac{10}{11})^2=(-\frac{10}{11})(-\frac{10}{11})=\frac{100}{121}\)

 

sin2 θ  +  100/121   =  1       Subtract   100 / 121  from both sides.

 

sin2 θ   =   21/121                Since sin is negative in Quad III, take the negative sqrt of both sides.

 

sin θ     =  \(-\sqrt{\frac{21}{121}}\)

 

sin θ     =   \(-\frac{\sqrt{21}}{11}\)

hectictar  Oct 10, 2017
edited by hectictar  Oct 10, 2017

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