find the six trigonometric function

Question

Find the six trigonometric function????

Find the value of the six trigonometric functions if the terminal side is on the line in the specified quadrant y=1/3 x in quadrant 3 one point on that line is (-6,-2)

  

Please chec k my answer

cos(theta)=-6√40=-12√10

sin(theta)=-2√40=-4√10

tan(theta)=-2/-6=1/3

csc(theta)=1/-4√10

sec(theta)=1/-12√10

cot(theta)=1/1/3

Answer 1

You are on the right track, Daisy, but need a little cleaning up. We know that at a point (x,y) on the coordinate plane, where r is the radius or the distance from the point to the origin, that

 

cos θ = x/r

sin θ = y/r

tan θ = y/x

csc θ = r/y

sec θ = r/x

cot θ = x/y

 

It looks like you have correctly calculated the radius to be √40. However, when you worked out some of the ratios, you missed the fraction.

 

cos θ would be -6/√40. If we rationalize this (-6√40)/40 = (-3√10)/10

Answer 2

Daisy, you may have mis-copied from the last time we did this problem.

Your idea was right.

 

cos(theta)=-6/√40=-6/2√10= -3/√10    (You lost the "/")

sometimes your teacher may want you to move the radical from the denominator to the numerator.

-3/√10 = -3√10 / 10   
sin(theta)=-2/√40=-2/2√10 = -1/√10    (You lost the "/")
tan(theta)=-2/-6=1/3           <-- tan theta is correct!

1 more comment. 1/1/3 = 3

 

sec(theta) = -√(10)/ 3

csc(theta) = -√10

cot(theta) = 3