Question

how does tan(5pi/4) = 1
how does tan(5pi/4) = 1
@Mathematics

Answer 1

use the unit circle

Answer 2

a full rotation is 2pi

Answer 3

i still dont understand sry

Answer 4

\[\sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}\]\[\cos\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}\]
\[\tan\frac{5\pi}{4}=\frac{\sin\frac{5\pi}{4}}{\cos\frac{5\pi}{4}}=1\]

Answer 5

ok thanks joemath

Answer 6

\[\tan(\frac{5\pi}{4})\] is like \[\frac{\pi}{4} * 5\] so you can just evaluate \[\tan(\frac{\pi}{4})\] and then see what quadrant 5pi/4 is in and then figure out if tan is positive or negative in that quadrant.

Answer 7

I strongly recommend you memorize the points to \[\frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{6}\] because if you know those three then you know any point of any angle.

Answer 8

alright, i just jot those down, how do i use them to find any point of any angle

Answer 9

dont forget:\[\frac{\pi}{2}\]No love for the right angles :(
lol

Answer 10

lol yeah.. \[\frac{\pi}{2}, \pi , \frac{3\pi}{2}\] are also good to memorize

Answer 11

and \[2\pi\]

Answer 12

but say you want to evaluate \[\sin(\frac{2\pi}{3})\]. This is the same as \[2 * \frac{\pi}{3}\] so just evaluate \[\sin(\frac{\pi}{3})\] which you have memorized. Then figure out what quadrant \[\frac{2\pi}{3}\] is in and then figure out if sin is positive or negative in that quadrant.

Answer 13

so just by memorizing the points to the angles 30, 45, and 60 you can solve anything

Answer 14

oh ok

Answer 15

hey tyler im glad u came around again...i had two other questions

Answer 16

first how does the 1st positive two asymptote of y= cot(2x - pi/2) = pi/4 and 3pi/4

Answer 17

ok.. first you have to know that \[\cot(x) = \frac{1}{\tan(x)}\]

Answer 18

now in order for a function to have a asymptote, the denominator must equal 0.. So we need to find where tangent is 0

Answer 19

but the unit circle only has cos and sin...how do i determine tan

Answer 20

\[\tan(x) = \frac{\sin(x)}{\cos(x)}\]

Answer 21

so tan would be 0 at 0 and pi

Answer 22

yeah thats what i was thinking because everything else on the unit circle, if you set it at sin/cos the minimum is like 1

Answer 23

so using 0 and pi...how do get to pi/4 and 3pi/4

Answer 24

right.. now you factor out a 2 out of the parenthesis\[y = \cot(x - \frac{\pi}{4})\] to get a phase shift of + pi/4

Answer 25

so the standard cot(x) functions has a asymptote at 0 and pi but its shifted to the right by pi/4 units.. so add pi/4 to 0 and pi.. but i get pi/4 and 5pi/4.. hmmm

Answer 26

this is what i have...
tanu is undefined when cosu = 0.
The first positive number where cosu = 0 is u = pi/2.
Solve 2x + pi = pi/2 to find the vertical asymptote.
2x + pi = pi/2
2x = -pi/2
x = -pi/4
-pi/4 is negative of course so try the next number where cosu = 0, 3pi/2.
2x + pi = 3pi/2
2x = pi/2
x = pi/4
You want a second positive asymptote so solve for 5pi/2.
2x + pi = 5pi/2
2x = 3pi/2
x = 3pi/4
The first two positive asymptotes are x = pi/4 and x = 3pi/4.

Answer 27

hmm you got the right answer.. i dunno where i messed up at :O

Answer 28

maybe i am going about it wrong.

Answer 29

ok thats fine...i can email my teacher about that...also

how do you determine when sin=1/3 cos= 2/sqrt2

i dont see 1/3 on the unit circle

Answer 30

There isnt a 1/3 on a unit circle so you have to use a right triangle.

Answer 31

wait hold on.. lol

Answer 32

i made a huge mistake lol