Trigonometric function question?

sin x = -(4/5)

tan x <0

cos x =+-sqrt(1-sin^2 x)

As tan x <0, sin x < 0 and tan x=sin x/cos x,

then cos x>0

cos x=sqrt(1-(16/25))=

=sqrt(9/25)=3/5

b) sin -(x)=-sin(x)=-(-4/5)=4/5

c)csc x =1/sin x=1/(-4/5)=-5/4

You have to start with the unit circle and the basic trig identity.

Remember to view these problems as a circle centered at (0,0) with a radius of 1.

sin^2 + cos^2 = 1

start at the point (1,0) which has an angle of zero

As you travel counterclockwise, your angle increases. The length of y-value is the sin component and the length of the x-component is the cos component.

For a tangent to be negative, the x and y values have to have opposite signs, so that would mean you are in the second or fourth quadrants. A negative sin means you are in the third or fourth quadrant.

SO, putting that together, you are in the fourth quadrant.

So, cos x would have to be a positive 3/5. (4/5)^2 + (3/5)^2 = 1

For b, the sin of the opposite of x puts you in the first quadrant, so the sin is a positive 4/5

For c, csc is the reciprocal of sin, so 1/(-4/5) = -5/4

Start here. We know that sin is the relationship of the angle's opposite side to the hypotenuse. If hypot. is 5, then Pythagorus tells us the third side is 3, or (-3). Draw a triangle, starting at (0,0). Add a pont at (-4,0) and the third point? Either at (0,3) or (0,-3). See if you can figure the rest out.