x=rcos thetax=rcosθ
x=7*cos((-7pi)/12)x=7⋅cos(−7π12)
if we add 1 revolution =2pi=2π to the angle then
2pi+(-7pi)/12=(17pi)/122π+−7π12=17π12
x=7*cos((17pi)/12)x=7⋅cos(17π12)
using double angle formulas
x=7*cos((17pi)/12)=7*cos(pi+(5pi)/12) x=7⋅cos(17π12)=7⋅cos(π+5π12)
x=7*cos(pi+(5pi)/12)x=7⋅cos(π+5π12)
x=7[cos pi *cos ((5pi)/12)- sin pi *sin ((5pi)/12)] x=7[cosπ⋅cos(5π12)−sinπ⋅sin(5π12)]
x=7*(-cos((5pi)/12)-0)x=7⋅(−cos(5π12)−0)
x=7*(-1)*cos ((5pi)/12)x=7⋅(−1)⋅cos(5π12)
Using double angle again
x=-7*cos(pi/3+pi/12)x=−7⋅cos(π3+π12)
x=-7*[cos (pi/3) *cos (pi/12)-sin (pi/3) *sin (pi/12)]x=−7⋅[cos(π3)⋅cos(π12)−sin(π3)⋅sin(π12)]
Recall the special angles 60^@=pi/360∘=π3 and 30^@=pi/630∘=π6 and pi/12=1/2*pi/6π12=12⋅π6 Use Half-Angle formulas for functions of pi/12π12
cos (pi/12)=sqrt((1+cos pi/6)/2)=sqrt((1+sqrt3/2)/2)=1/2sqrt(2+sqrt3)cos(π12)=√1+cosπ62=√1+√322=12√2+√3
sin (pi/12)=sqrt((1-cos pi/6)/2)=sqrt((1-sqrt3/2)/2)=1/2sqrt(2-sqrt3)sin(π12)=√1−cosπ62=√1−√322=12√2−√3
so that
x=-7*[cos (pi/3) *cos (pi/12)-sin (pi/3) *sin (pi/12)]x=−7⋅[cos(π3)⋅cos(π12)−sin(π3)⋅sin(π12)]
becomes
x=-7*[1/2 *(sqrt(2+sqrt3))/2-sqrt3/2 *sqrt(2-sqrt3)/2]x=−7⋅[12⋅√2+√32−√32⋅√2−√32]
x=7/4*(sqrt(6-3sqrt3)-sqrt(2+sqrt3))=-1.81173x=74⋅(√6−3√3−√2+√3)=−1.81173
Do the same for yy and come up with
y=-7/4*(sqrt(6+3sqrt3)+sqrt(2-sqrt3))=-6.76148y=−74⋅(√6+3√3+√2−√3)=−6.76148
