Help evaluating tan(pi/16)?

Question

Can someone help show me how to use half-angle identities to evaluate tan(pi/16)? Thanks.

sorry forgot to mention: i need to keep it in radical form

Answer 1

I know that the tan(x) = sin(x)/cos(x)

I know sin(π/4) = cos(π/4) = √(2/4)

I also know that cos(a+b)= cos(a)cos(b) - sin(a)sin(b)
if a=b, then I can deduce that
cos(2a) = cos²(a) - sin²(a)

Since I know hat sin²(a) + cos²(a) = 1, I can deduce that
cos(2a) = 2cos²(a) - 1 From this I derive
cos(a) = √[(1 + cos(2a))/2

Or, if you prefer
cos(a/2) = √[(1 + cos(a))/2]

But we want to find tan(a/4) so we have to go through the process one more time
cos(a/4) = √[(1 + √[(1 + cos(a))/2])/2
sin(a/4) = √[(1 - √[(1 + cos(a))/2])/2]

Since tan(a/4) = sin(a/4)/cos(a/4) you now have
Tan(π/16) = √[(1 - √[(1 + cos(π/4))/2])/2] / √[(1 + √[(1 + cos(π/4))/2])/2

Now, if your teacher wants it in terms of tan rather than cos, sorry. I'm try to keep the stuff to memorize down to a minimum.... I only remember how to derive the sine and cosine of the sums of angles.

Answer 2

An identity you can use is

tan(x/2) = [1 - cos(x)]/sin(x)

if you know the exact values of cos(pi/8) and sin(pi/8).

I don't either. So use

cos²(x/2) = (1+cos(x))/2 and sin²(x/2) = (1-cos(x))/2 with x = pi/4

Since all these angles are in Quadrant I, you can use

cos(x/2) = sqrt[(1+cos(x))/2] and sin(x/2) = sqrt[(1-cos(x))/2]

I get (after a lot of arithmetic) sqrt(4 + 2sqrt(2)) - sqrt(2) - 1

Answer 3

Have to think about that one, but I would use the Taylor series instead, much easier for numerical approximations. The Taylor series for tangent x + (x^3)/3...Just plug in your number and go. Usually in the Taylor series you only need to keep one to two terms, so tan (pi/16) approximately equals pi/16 to good approximation.