Question

How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle `pi/12`?

Answer 1

`1/2sqrt(2-sqrt3),1/2sqrt(2+sqrt3),sqrt(7-4sqrt3)`

Explanation:

`"since "pi/12" is in the first quadrant all of the trig. ratios"`
`"will be positive"`

`•color(white)(x)sin(x/2)=+-sqrt((1-cosx)/2)`

`"here "(x/2)=pi/12rArrx=pi/6`

`sin(pi/12)=+sqrt((1-cos(pi/6))/2`

`color(white)(xxxxxx)=sqrt((1-sqrt3/2)/2`

`color(white)(xxxxxx)=sqrt(((2-sqrt3))/4)=1/2sqrt(2-sqrt3)`
`color(blue)"---------------------------------------------------------"`

`•color(white)(x)cos(x/2)=+-sqrt((1+cosx)/2)`

`cos(pi/12)=+sqrt((1+cos(pi/6))/2)`

`color(white)(xxxxxx)=sqrt((2+sqrt3)/4)=1/2sqrt(2+sqrt3)`
`color(blue)"----------------------------------------------------------"`

`•color(white)(x)tan(x/2)=+-sqrt((1-cosx)/(1+cosx))`

`tan(pi/12)=+sqrt((2-sqrt3)/(2+sqrt3))`

`"multiply the numerator/denominator by "(2-sqrt3)`

`rArrtan(pi/12)=sqrt(7-4sqrt3)`