If sinx=−4/5 and cotx<0, then find cos2x

Question

Find the exact functional value.

Answer 1

Since sine < 0 and cot <0 then the cos(x) will be in Quadrant 4. This is a 3-4-5 triangle, so cos(x) = 3/5

cos(2x) = 2cos²(x) - 1 = 2(9/25) = -7/25

Answer 2

x is in quadrant III

cotx>0 in quadrant III and I

sinx<0 in quadrants III and IV

only in quadrant III is sinx<0 and tanx>0

if sinx=-4/5, then cosx=-3/5. It's a 3,4, 5 right triangle

then cos2x = cos^2x - sin^2x = 9/25 - 16/25 = -7/25

2x is in quadrant II, it's a 7-24-25 right triangle

Answer 3

cos2x = 1 - 2sinx^2

sinx = -4/5 so

cos2x = 1 - 2(-4/5)^2 = 1 - 32/25 = -7/25

Answer 4

Sine and tangent are both negative, so that's quadrant 4.

Sine is-4/5, so opposite is -4 and hypotenuse is 5.

Then adjacent=-3 by pythagorean

So then cosine is -3/5.

Cos 2x= cos^2 - sin^2

= (-3/5)^2-(-4/5)^2

= 9/25-16/25

=-7/25