# math help trig identities?

prove the following identity

cos (pi/12) = the sqroot of ( 2 +sqrt 3) / 2

Steps (not all steps):

1. cos (pi/6/2)

2. plus or minus the sqroot of 1+cos(pi/6)/2

my question is how the heck did you get the radical and 1+cos(pi/6) from just cos (pi/6/2)?

where did all this come from.

Relevance
• Anonymous
8 years ago

pi/12 = 15 degrees

cos (15) = cos (45 - 30) = cos 45 cos 30 + sin 45 sin 30

=(1/sqrt(2)) (sqrt(3)/2) + (1/sqrt(2)) (1/2)

=(1+sqrt(3)) /2sqrt(2)

You can also use the half angle formula

cos(x/2) = sqrt( (1+cos x)/2)

So cos (15)=cos(30/2) = sqrt (( 1+ cos 30)/2) = sqrt (( 1 + sqrt(3)/2)/2) = sqrt(2+sqrt(3))/2

This square root simplifies to the answer above (simply square the answer above to confirm).

• It's using a half angle identity.

cos(t/2) = sqrt((1+cos(t))/2)

cos(pi/12) = cos((pi/16)/2)

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

Use the 30-60-90 triangle to find the value of cos(pi/6). The sides of the triangle are 1/2, 2/2 and sqrt(3)/2. So cos(pi/6) is sqrt(3)/2

cos(pi/12) = sqrt((1+sqrt(3)/2)/2)

cos(pi/12) = sqrt(1/2+sqrt(3)/4)

cos(pi/12) = sqrt(2/4+sqrt(3)/4)

cos(pi/12) = sqrt((2+sqrt(3))/4)

cos(pi/12) = sqrt(2+sqrt(3))sqrt(1/4)

cos(pi/12) = sqrt(2+sqrt(3))/2