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Let cos(−θ) = 4 / 5 and tanθ > 0.

 

What is the value of sin(−θ)?

 

4 / 3

−4 / 5

4 / 5

−3 / 5

GuestMember  Feb 28, 2018
 #1
avatar+7155 
+2

By the Pythagorean identity...

 

( sin(-θ) )2 + ( cos(-θ) )2  =  1

                                                 We are given that  cos(-θ) = 4/5

( sin(-θ) )2 + ( 4/5 )2  =  1

 

( sin(-θ) )2 + 16/25  =  1

                                                 Subtract  16/25  from both sides of the equation.

( sin(-θ) )2  =  1 - 16/25

 

( sin(-θ) )2  =  9/25

                                                 Take the  ±  square root of both sides.

sin(-θ)  =  ±√[ 9/25 ]

 

sin(-θ)  =  ± 3 / 5

 

Since cos(-θ) is positive,  -θ  must be in Quadrant I or Quadrant IV.

 

Since tan θ > 0 ,  θ  must be in Quadrant I or III. That means  -θ  must be in Quadrant IV or II .

 

So  -θ  has to be in Quadrant IV, and  sin(-θ) has to be negative.

 

sin(-θ)  =  - 3 / 5

hectictar  Mar 1, 2018

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