SOLUTION: **Find the exact value of sec (71[pi]/12)** the answer is (root 6 - root 2) but i don't know how to get to the answer I need the steps and solution ASAP!!

Click here to see ALL problems on Trigonometry-basics

Question 535813: **Find the exact value of sec (71[pi]/12)**
the answer is (root 6 - root 2) but i don't know how to get to the answer
I need the steps and solution ASAP!!
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website!
Find the exact value of sec (71[pi]/12)
the answer is (root 6 - root 2) but i don't know how to get to the answer
**
Finding the reference angle:
Notice that the given angle is π/12 short of 72π/12 or 6π, the origin; so the reference angle is π/12 in quadrant IV where sec and cos>0 and sin<0
sec(π/12)=1/cos(π/12)
cos(π/12)=cos(π/3-π/4)=cos(4π/12-3π/12)
Using cos addition formula: Cos(s-t)=cos s cos t+sin s sin t
cos(π/3+π/4)=cos(π/3)cos(π/4)+sin(π/3)sin(π/4)
cos(π/3+π/4)=1/2*√2/2+√3/2*√2/2)=√2/4+√6/4=(√2+√6)/4)
cos(π/3+π/4)=(√2+√6)/4)
Take reciprocal
sec(π/3+π/4)=4/(√2+√6)
rationalize denominator of 4/(√2+√6)
multiply numerator and denominator by √2-√6
4(√2-√6)/(√2+√6)(√2-√6) (makes denominator difference of squares)
4(√2-√6)/(2-6)
4(√2-√6)/-4
√6-√2
Ans:
sec (71[pi]/12)=√6-√2