Question

Trig/Precalc Question:
tan(pi/16)
Use the half-angle formulas to come up with the an exact expression for the function value above.

Answer 1

Do you know how to start this problem?

Answer 2

well i know well i know that the half angle formula is
\[\tan^2x=(1−\cos(2x))/(1+\cos(2x))\]

Answer 3

Well right now you have a tan(pi/2^{a}) and the you are trying to make it so the function equal something that has to do with sin(pi) and cosine(pi) with no coefficients that you can cancel out correct?
so tan(pi/16) = sin(pi/8) / (1+cos(pi/8))

And then you just keep on breaking sin and cos until you get a even pi and then it is just plugging in 0 and pi, which should be easy enough.

Answer 4

so would 1/ sqrt2 be the answer?

Answer 5

Erm, I'm not finished yet, sorry.

Answer 6

No I did not get that answer, mine was pretty complex,
What was your end result before you solved for the 0's and pi's?

Answer 7

I broke tan((pi)/16) into sin((pi)/8) / (1+cos((pi)/8))
.
Then sin((pi)/8) into sqrt((1-cos (pi)/4)/2)
And 1 + cos((pi)/8) into 1+ sqrt((1+cos (pi)/4)/2)

Then cos (pi/4) = sqrt((1+cos (pi)/2)/2)

Then cos (pi/2) = sqrt((1+cos (pi))/2)

Then substituted them back into the previous equation until I got a really big equation, Then set cos pi = -1 and simplified that big equation.

Answer 8

The answer that I got was
\[\sqrt{0.5 * ((1+\sqrt{2})/\sqrt{2})}/(1+\sqrt{0.5 * ((1+\sqrt{2})/\sqrt{2})})\]