use a sum or difference identity to find an exact value of sin 2pi over 12

Sagot :

5pi/12 = 3pi/12 + 2pi/12 = pi/4 + pi/6 

sin(a + b) => sin(a)cos(b) + sin(b)cos(a) 
cos(a + b) => cos(a)cos(b) - sin(a)sin(b) 

sin(5pi/12) => 
sin(3pi/12 + 2pi/12) => 
sin(pi/4 + pi/6) => 
sin(pi/4)cos(pi/6) + sin(pi/6)cos(pi/4) => 
(sqrt(2)/2) * (sqrt(3)/2) + (1/2) * (sqrt(2)/2) => 
(sqrt(2)/4) * (1 + sqrt(3)) 

cos(5pi/12) => 
cos(pi/4 + pi/6) => 
cos(pi/4)cos(pi/6) - sin(pi/4)sin(pi/6) => 
(sqrt(2)/2) * (sqrt(3)/2) - (sqrt(2)/2) * (1/2) => 
(sqrt(2)/4) * (sqrt(3) - 1) 

tan(5pi/12) => 
sin(5pi/12) / cos(5pi/12) => 
(sqrt(2)/4) * (sqrt(3) + 1) / ((sqrt(2)/4) * (sqrt(3) - 1)) => 
(sqrt(3) + 1) / (sqrt(3) - 1) => 
(sqrt(3) + 1)^2 / (3 - 1) => 
(3 + 2 * sqrt(3) + 1) / 2 => 
(4 + 2 * sqrt(3)) / 2 => 
2 + sqrt(3) 

csc(5pi/12) => 
1 / sin(5pi/12) => 
4 / (sqrt(2) * (sqrt(3) + 1)) => 
4 * sqrt(2) * (sqrt(3) - 1)) / (2 * (3 - 1)) => 
4 * sqrt(2) * (sqrt(3) - 1)) / (2 * 2) => 
sqrt(2) * (sqrt(3) - 1) 

sec(5pi/12) => 
1/cos(5pi/12) => 
4 / (sqrt(2) * (sqrt(3) - 1)) => 
4 * sqrt(2) * (sqrt(3) + 1)) / (2 * (3 - 1)) => 
sqrt(2) * (sqrt(3) + 1) 

cot(5pi/12) => 
1/tan(5pi/12) => 
1 / (2 + sqrt(3)) => 
(2 - sqrt(3)) / (4 - 3) => 
2 - sqrt(3)