Question

Find tan(3pi/8) using half-angle identities. !!!!!!

Answer 1

\[\frac{3\pi}{8}\] is half of
\[\frac{3\pi}{4}\] so use
\[\tan(\frac{3\pi}{8})=\frac{\sin(\frac{3\pi}{2})}{1+\cos(\frac{3\pi}{2})}\]

Answer 2

damn typo, sorry. should be
\[\tan(\frac{3\pi}{8})=\frac{\sin(\frac{3\pi}{4})}{1+\cos(\frac{3\pi}{4})}\]

Answer 3

no i do know that it's simplying the answer that i'm having trouble with i guess

Answer 4

like that's [(sqrt2)/2]/[1-(sqrt2)/2]... what do I do from there.... oy

Answer 5

If I do it using the other formula (1-cosx)/sinx I get [1+(sqrt2)/2]/[(sqrt2)/2] = [(2+sqrt2)/2]*[2/sqrt2] = [2+sqrt2]/[sqrt2] ... right? That is accurate?

Answer 6

Then from there to rationalize the fraction I multiply both num and denom by sqrt2 ... giving me... [2(sqrt2)+2]/[2], which = sqrt2 + 1.

Answer 7

Alrighty....