Question

PRECAL! Is this correct about tan(π/16)? I know it shouldn't look exactly like this in one straight line...

Answer 1

I know its really just a lot of plugging in... \[\tan \frac{ π }{ 16 }= \frac{ \sqrt{(1-\cosπ/8)} }{ \sqrt{(1+\cosπ/8)} } \]

Answer 2

tan = sin/cos

Answer 3

im not aware of any nice ways to address pi/16 tho.....

Answer 4

\[\tan \frac{ π }{ 16 } = \frac{ \sin \frac{ π }{ 8 }}{ 1+\cos \frac{ π }{ 8 } }\]

Answer 5

Is this the approach you were thinking of?

Answer 6

\[tan(x)=\frac{sin(x)}{cos(x)}\]
was the approach i was thinking of; but then there are half anlge and double anlge formulas that compound the issue

Answer 7

Says to "use the half-angle formulas to come up with an exact expression..."

Answer 8

the nice angles tend to be those constructed on the unit circle: 0,pi/3, pi/4, pi/6, and pi/2

trying to work these out into pi/16 can be a pain to say the least

Answer 9

tan(π/16) = sin(π/8)/(1+cos(π/8)--------------------(1)

Similarly, applying same formula to (π/4)
tan(π/8) = sin(π/4)/(1+cos(π/4) = 1/(√2 +1) = √2 -1
sec² (π/8) = 2 +1-2√2 +1 = 2*(2-√2) or
cos(π/8) = 1/√[2*(2-√2)] =[(1/4)*√{2*(2+√2)}]---------(2) and
sin (π/8) =[{(√2 -1)/4}*√{2*(2+√2)}] ------------------(3)

Substituting from (2) and (3) in (1), we get
tan(π/16) =[{(√2 -1)/4}*√{2*(2+√2)}]/[1+[(1/4)*√{2*(2+√2)…
tan(π/16) =[{(√2 -1)}*√{2*(2+√2)}]/[4+{√{2*(2+√2)}]

Answer 10

i cant focus on these ....

Answer 11

\[tan(\beta)=tan(\frac12\alpha)=\frac{tan(\alpha)}{1+sec(\alpha)}\]
\[\frac{tan(\alpha)}{1+\sqrt{tan^2(\alpha)+1~}}\]
\[tan(\alpha)=tan(\frac12\gamma)=\frac{tan(\gamma)}{1+sec(\gamma)}\]
let \(\gamma\)=pi/4

Answer 12

\[tan(\alpha) = \frac1{1+\sqrt2}\]
\[tan^2(\alpha) = \frac1{3+2\sqrt2}\]
\[tan(\beta)=\frac{1}{1+\sqrt{\cfrac1{3+2\sqrt2}+1~}}\]
\[\frac{1}{1+\sqrt{\cfrac{1+3+2\sqrt2}{3+2\sqrt2}~}}\]
\[\frac{1}{1+\sqrt{\cfrac{4+2\sqrt2}{3+2\sqrt2}~}}\]
maybe ....

Answer 13

okay... That was a different method