Find the exact value tan(9pi/8) Using half life formula
Would the 9pi/8 change to 9pi/4?
Hehe, I like how you called it half-life formula :P Well, the idea is that \[\tan(x/2) = \frac{ 1-cosx }{ sinx }\] What you want to do is rewrite 9pi/8 as x/2. So if 9pi/8 = x/2, what is x?
Oh you already answered that, yes, you would use 9pi/4 in the formula, lol.
I meant half angle lol, my formula is different from that one
\[\tan \theta/2=\sqrt{1-\cos \theta/1+\cos \theta}\]
It's an equivalent formula. There are 3 formulas you can usefor tan(x/2), i just chose one of the simple ones. These are the 3 you can use: tan(x/2) = \[\sqrt{\frac{ 1-cosx }{ 1+cosx }}\] \[\frac{ 1-cosx }{ sinx }\] or \[\frac{ sinx }{ 1+cosx }\]
Oh okay
Yep. So choose whichever one you like and plug in 9pi/4 :)
But 9pi/4 isnt on my unit circle
Well, any value on the unit circle is equivalent to adding or subtracting multiples of 2pi. So what we can do is subtract 2pi from 9pi/4 to get an equivalent angle that wil be on the unit circle.
is it just simply 7pi/4?
Well, 2pi is equivalent to 8pi/4. Subtracting that for 9pi/4, we would have 9pi/4 - 8pi/4 = pi/4. Does that make sense?
I think so yeah, how did you know what 2pi is equivalent to 8pi/4
\[\frac{ 2\pi }{ 1 }*\frac{ 4 }{ 4 } = \frac{ 8\pi }{ 4 }\]
\[=\sqrt{1-(2/\sqrt2)/1+(2/\sqrt2)}\] Kay, With my formula I have this
You have the square roots reverse. it should be \(\sqrt{2}/2\) on each of those.
Oh right
Would I then multiple the 2 to everything?
To reduce it, yes :3
Should I leave it like it is after that
That would be up to your professor. You would have: \[\sqrt{\frac{ 2-\sqrt{2} }{ 2+\sqrt{2} }}\] Now, because of the formula I mentioned, this is equivalent to: \[\frac{ \sqrt{2} }{ 2+\sqrt{2} }\] which could be rationalized. So considering the two are equal, it really is up to the professor on how much crazy simplification you have to do.
thank you, I have two more problems to do
Thats fine, seems like I have time
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