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Mathematics 7 Online
OpenStudy (sleepyjess):

Find the exact value by using a half-angle identity. \(tan\dfrac{7\pi}{8}\)

OpenStudy (anonymous):

ok tan(pie\8)=1-cos(pie/4) /1-sin(pie/4) thats the half angle identity right so we know now that cos(pie/4) =root2/2 similarly to sine (pie/4) so the you solve :)

OpenStudy (anonymous):

oh sorry im really tired i messed up sorry the identity is tan(pie/8)=1-cos(pie/4) / 1+cos(pie/4)

OpenStudy (anonymous):

haha we meet again JESS

OpenStudy (anonymous):

easy......which half angle formula do u want??

OpenStudy (sleepyjess):

Ummmm I don't know. I have no clue on how to even start this.

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

so I am guessing without any conversions!!

OpenStudy (anonymous):

OK here we go

OpenStudy (anonymous):

where do u think 7pi/8 lies

OpenStudy (anonymous):

Is it b/w (0,pi) or (pi,2pi)??

OpenStudy (anonymous):

cmon girl......we're losing daylight !!

OpenStudy (sleepyjess):

I think pi, 2pi

OpenStudy (anonymous):

@sleepyjess

OpenStudy (anonymous):

seriously??

OpenStudy (anonymous):

when I say think I meant find the value not guess!!!

OpenStudy (anonymous):

wake up!!

OpenStudy (jhannybean):

You can use \[\tan\left(\frac{x}{2}\right) = \frac{1-\cos(x)}{\sin(x)}\]\[\tan\left(\dfrac{7\pi}{8}\right) = \tan\left(\dfrac{\dfrac{7\pi}{4}}{2}\right)\]

OpenStudy (anonymous):

NO @Jhannybean LET HER WAKE UP FIRST

OpenStudy (anonymous):

there's a far better way dude

OpenStudy (anonymous):

easier...

OpenStudy (jhannybean):

\[\tan\left(\dfrac{\color{red}{\dfrac{7\pi}{4}}}{2}\right) =\tan\left(\frac{\color{red}x}{2}\right)\]

OpenStudy (anonymous):

still easier

OpenStudy (jhannybean):

Therefore, \[\tan\left(\frac{\color{red}x}{2}\right) = \frac{1-\cos(\color{red}x)}{\sin(\color{red}x)}\]

OpenStudy (jhannybean):

Do you understand, @sleepyjess ?

OpenStudy (sleepyjess):

Kind of

OpenStudy (jhannybean):

Follow my posts please.

OpenStudy (anonymous):

dude do u guys dont want easier versions??

OpenStudy (anonymous):

meh...fine!!

OpenStudy (jhannybean):

Let the person learn it however he/she likes, stop enforcing your methods, please.

OpenStudy (jhannybean):

Go ahead and show her then, and stop spamming the question :)

OpenStudy (sleepyjess):

So you just do tan(whatever) over 2?

OpenStudy (jhannybean):

Pretty much, there are 3 different identities for the half angle for tan. You can use whatever one you like

OpenStudy (jhannybean):

\[\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}\]\[\tan\left(\frac{x}{2}\right) = \frac{1-\cos(x)}{\sin(x)}\]\[\tan\left(\frac{x}{2}\right) = \frac{\sin(x)}{1+\cos(x)}\]

OpenStudy (sleepyjess):

So, \(\sf \tan\left(\dfrac{\color{red}{\dfrac{7\pi}{4}}}{2}\right) = \sqrt{\dfrac{1-cos\dfrac{7\pi}{4}}{1+cos\dfrac{7\pi}{4}}}\)

OpenStudy (jhannybean):

I wasn't really comfortable with the square root, so I chose to use the second form instead, lol.

OpenStudy (sleepyjess):

oh lol that was a lot to type out haha

OpenStudy (jhannybean):

:) First of all, do you understand why \(\dfrac{7\pi}{8} \longrightarrow \dfrac{7\pi}{4}\) ?

OpenStudy (sleepyjess):

so, \(\sf \dfrac{1-cos\dfrac{7\pi}{4}}{sin\dfrac{7\pi}{4}}\)

OpenStudy (anonymous):

ya I get ur method @Jhannybean ..LOL

OpenStudy (sleepyjess):

Because 8/4 is 2 right?

OpenStudy (anonymous):

this method is useful for people with calculators

OpenStudy (jhannybean):

Not exactly. \[\frac{7\pi}{8} =\dfrac{\color{red}{\dfrac{7\pi}{4}}}{2}\]

OpenStudy (sleepyjess):

Correction, because 8/2 is 4

OpenStudy (anonymous):

lets say we change it to something simpler

OpenStudy (anonymous):

pi>7pi/8>0

OpenStudy (anonymous):

lies in second quadrant

OpenStudy (jhannybean):

Correct.

OpenStudy (anonymous):

so tan is negative

OpenStudy (anonymous):

now......tan 7pi/8=tan(pi-pi/8)

OpenStudy (jhannybean):

|dw:1419983753548:dw|

OpenStudy (anonymous):

and we knw tan (pi-x)=-tanx

OpenStudy (anonymous):

u with me gurl??

OpenStudy (sleepyjess):

Yeah

OpenStudy (anonymous):

so then we get -tan pi/8

OpenStudy (anonymous):

now we folloy @Jhannybean

OpenStudy (anonymous):

so we get pi/4

OpenStudy (anonymous):

using \[\tan 2\theta = 2 \tan \theta/1-\tan^2\theta\]

OpenStudy (anonymous):

U know this formula rt??

OpenStudy (sleepyjess):

Yes, it is in my lesson

OpenStudy (anonymous):

so know we change it

OpenStudy (anonymous):

in terms of tan theta

OpenStudy (anonymous):

or follow @Jhannybean whichever u get

OpenStudy (jhannybean):

Go ahead an finish up @cambrige , writing up both explanations will only confuse the person asking.

OpenStudy (anonymous):

@Jhannybean thinks I think he is incompetent to solve problems.Fine I quit..Solve it for her

OpenStudy (anonymous):

and stop pmming me @Jhannybean

OpenStudy (anonymous):

ok so,...yes we were at -tan pi/8

OpenStudy (anonymous):

on changing it for half angle

OpenStudy (anonymous):

then we will get formula

OpenStudy (anonymous):

tan x/2= sin x/1-cos x

OpenStudy (anonymous):

oi........U there???

OpenStudy (sleepyjess):

Yes

OpenStudy (anonymous):

after that U get ??

OpenStudy (anonymous):

taking x=pi/4

OpenStudy (anonymous):

tan (pi/4)/2=

OpenStudy (anonymous):

sin (pi/4)/1-cos pi/4

OpenStudy (anonymous):

mind telling me the value of cos and sin pi/4???

OpenStudy (anonymous):

okay.......Me waiting??

OpenStudy (anonymous):

@sleepyjess

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