How do you find the exact value of the six trigonometric functions of the angle whose terminal side passes through (2/3, 5/2)?

1 Answer
turksvids
Jan 11, 2018

cos(theta)=4/sqrt(241), sin(theta)=15/sqrt(241), and tan(theta)=15/4, sec(theta)=sqrt(241)/4, csc(theta)=sqrt(241)/15, and cot(theta)=4/15.

Explanation:

The points (2/3,5/2) and (4,15) fall on the same line and therefore the terminal side of the same angle, theta, and the second point is easier to work with. (I basically multiplied through by 6 to scale the point up.)

Calculate r=sqrt(4^2+15^2)=sqrt(16+225)=sqrt(241).

Knowing x, y, and r we can find:

cos(theta)=x/r, sin(theta)=y/r, and tan(theta)=y/x.

For this problem we have:

cos(theta)=4/sqrt(241), sin(theta)=15/sqrt(241), and tan(theta)=15/4.

The remaining functions are found by taking reciprocals:

sec(theta)=sqrt(241)/4, csc(theta)=sqrt(241)/15, and cot(theta)=4/15.

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