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**+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