Question

How do you give the six trigonometric function values of pi/3?

Answer 1

`"see explanation"`

Explanation:

`"using the right triangle with angles and sides"`

`pi/2,pi/6,pi/3larrcolor(red)" angles"`

`1,sqrt3,2larrcolor(red)" sides"`

`"in relation to "cos(pi/3)`

`1" is adjacent ",sqrt3" is opposite",2" is hypotenuse"`

`•color(white)(x)cos(pi/3)=1/2`

`•color(white)(x)sec(pi/3)=1/cos(pi/3)=1/(1/2)=2`

`•color(white)(x)sin(pi/3)=sqrt3/2`

`•color(white)(x)csc(pi/3)=1/sin(pi/3)=1/(sqrt3/2)=2/sqrt3`

`•color(white)(x)tan(pi/3)=sin(pi/3)/(cos(pi/3))=(sqrt3/2)/(1/2)=sqrt3`

`•color(white)(x)cot(pi/3)=1/tan(pi/3)=1/sqrt3`