Question

find the exact value of cot(9pi/8)

Answer 1

@jim_thompson5910 ?

Answer 2

i know this requires the half angle formula

Answer 3

first I would try to find tan(9pi/8)

Answer 4

\[1-\cos \theta/\sin\]

Answer 5

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

look on page 2 where it says "Half Angle Formulas "

Answer 6

oh isn't my formula correct?

Answer 7

I would use
\[\Large \tan\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1-\cos(\theta)}{1+\cos(\theta)}}\]

Answer 8

theta in this case is theta = 9pi/4

Answer 9

notice how

9pi/4 - 2pi = 9pi/4 - 8pi/4 = pi/4

so 9pi/4 and pi/4 are coterminal

Answer 10

so you could use theta = pi/4

Answer 11

ok and when it is 9pi/8 wouldn't you flip it and mult it by 2?

Answer 12

then plug that in the tan formula

Answer 13

focus on pi/4 for now. Plug that into the formula I gave you and tell me what you get

Answer 14

ok but when u subtracted by 2pi. where did 2pi come from

Answer 15

oh because coterminal?

Answer 16

yes adding or subtracting 2pi to any angle leads to another coterminal angle

Answer 17

is it \[\sqrt{4-\pi}\]

Answer 18

what is cos(pi/4) equal to?

Answer 19

\[\sqrt{2}/2\]

Answer 20

So you'd have
\[\Large \tan\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1-\cos(\theta)}{1+\cos(\theta)}}\]
\[\Large \tan\left(\frac{\frac{\pi}{4}}{2}\right) = \pm\sqrt{\frac{1-\cos\left(\frac{\pi}{4}\right)}{1+\cos\left(\frac{\pi}{4}\right)}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \pm\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}}\]

Answer 21

now simplify

Answer 22

\[\sqrt{3-2\sqrt{2}}\]

Answer 23

notice how using a calc, we get

tan(pi/8) = 0.4142135623731

so we can drop the minus in the plus/minus to get just this

\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}}\]

Answer 24

now let's simplify the right side

Answer 25

\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{\frac{2}{2}-\frac{\sqrt{2}}{2}}{\frac{2}{2}+\frac{\sqrt{2}}{2}}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{\frac{2-\sqrt{2}}{2}}{\frac{2+\sqrt{2}}{2}}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{2-\sqrt{2}}{2} \times \frac{2}{2+\sqrt{2}}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}\]
Hopefully this is making sense?

Answer 26

Now let's rationalize the denominator
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}\times\frac{2-\sqrt{2}}{2-\sqrt{2}}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{(2-\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{4-4\sqrt{2}+2}{(2)^2-(\sqrt{2})^2}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{4-4\sqrt{2}+2}{4-2}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{4-4\sqrt{2}+2}{2}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{\frac{6-4\sqrt{2}}{2}}\]
\[\Large \tan\left(\frac{\pi}{8}\right) = \sqrt{3-2\sqrt{2}}\]