Question

T is a real number and P=(square root of 3/2, 1/2) is the point on the unit circle that corresponds to t. How do you find the exact values of six trigonometric functions of t?

Answer 1

See explanation.

Explanation:

Given the point `P=(sqrt(3)/2;1/2)` on the angle's side we have:

`x=sqrt(3)/2`, `y=1/2`

`r=sqrt(x^2+y^2)=sqrt(3/4+1/4)=1`

Now we can calculate the trig. functions:

`sinT=y/r=1/2`

`cosT=x/r=sqrt(3)/2`

`tanT=y/x=1/2-:sqrt(3)/2=1/2xx2/sqrt(3)=1/sqrt(3)=sqrt(3)/3`

`cotT=x/y=1/tanT=sqrt(3)`

`secT=r/x=1-:sqrt(3)/2=2/sqrt(3)=(2sqrt(3))/3`

`cscT=r/y=1-:1/2=2`