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What is the value of cos(−θ)?

 Mar 4, 2018

Best Answer 

 #1
avatar+7354 
+5

By the Pythagorean identity...

 

sin2(-θ) + cos2(-θ)   =   1

                                             Plug in  (-3/5)  for  sin(-θ)

(-3/5)2 + cos2(-θ)   =   1

 

9/25   +  cos2(-θ)   =   1

                                             Subtract  9/25  from both sides of the equation.

cos2(-θ)   =   1  -  9/25

 

cos2(-θ)   =   16/25

                                             Take the  ±  square root of both sides.

cos(-θ)   =   ±√[ 16/25 ]

 

cos(-θ)   =   ± 4 / 5

 

Since  sin(-θ)  is negative,  -θ  must lie in Quadrant III or Quadrant IV.

Since  tan θ  is positive,  θ  must lie in Quadrant I or III, and so  -θ  must lie in Quadrant II or IV.

 

So  -θ  must lie in Quadrant IV, and  cos(-θ)  must be positive.

 

cos(-θ)   =   4 / 5

 Mar 4, 2018
 #1
avatar+7354 
+5
Best Answer

By the Pythagorean identity...

 

sin2(-θ) + cos2(-θ)   =   1

                                             Plug in  (-3/5)  for  sin(-θ)

(-3/5)2 + cos2(-θ)   =   1

 

9/25   +  cos2(-θ)   =   1

                                             Subtract  9/25  from both sides of the equation.

cos2(-θ)   =   1  -  9/25

 

cos2(-θ)   =   16/25

                                             Take the  ±  square root of both sides.

cos(-θ)   =   ±√[ 16/25 ]

 

cos(-θ)   =   ± 4 / 5

 

Since  sin(-θ)  is negative,  -θ  must lie in Quadrant III or Quadrant IV.

Since  tan θ  is positive,  θ  must lie in Quadrant I or III, and so  -θ  must lie in Quadrant II or IV.

 

So  -θ  must lie in Quadrant IV, and  cos(-θ)  must be positive.

 

cos(-θ)   =   4 / 5

hectictar Mar 4, 2018

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