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OK, look at the unit circle. FInd the angle in quadrant III with y=coordinate = -3/7. What's the x-coordinate? Just use the identity

cosΘ = ±√1-sin^2Θ. That's straight from the Pythagorean Theorem, since cosΘ is the x-coordinate, and sinΘ is the y-coordinate. x^2 + y^2 = 1, so x = ±√1-y^2.

So x = cosΘ = ±√1-9/49 = ±√40/49 = ±2/7√10

Do you want + or - ? Well, in Quadrant III, both coordinate are <0, so you want -

Now you have

sin = -3/7

cos = -(2/7)√10

And now you can get all the other trig functions from those.

I would draw an angle in quadrant III. Drop a perpendicular to the x axis.

Angle theta has sine = -3/7, and sine= opp/hyp or y/r.

Label the vertical leg -3, and the hypotenuse 7. Use Pythagorean theorem to find the horizontal leg, which would be -sqrt(40) , or -2sqrt(10)

Then use triangle ratios to find the other functions.

Cos(theta)= adjacent/hypot or x/r= -2sqrt(10)/7

Tan(theta)= opp/adj , or y/x= (-3)/ [-2sqrt(10)]

= 3/2sqrt(10) or 3sqrt(10)/20

Then use reciprocals for csc, sec, and cot.

Hoping this helps!

cos²θ = 1-sin²θ = 40/49

cosθ = -√(40/49) = -2√10/7

You can finish from here.

cscθ = 1/sinθ = -7/3