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\[\tan (\frac{ 17\pi }{ 12 })\]
how many 12's are in 17?
I know I have to use the equation \[\tan(a+b)=(\frac{ \tan(a)-\tan(b) }{ 1+\tan(a)\tan(b) })\]

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1
right so 17=12(1)+5 \[\tan(\frac{17}{12} \pi) \\ \tan([1+\frac{5}{12}]\pi) \\ \tan(\pi+\frac{5\pi}{12}) =\tan(\frac{5\pi}{12} ) \text{ since tangent function has period } \pi \] so now we are looking at tan(5pi/12) instead you said you hated the radian stuff so you can concert 5pi/12 to degrees if you want
\[\frac{5\pi}{12} \cdot \frac{180^o}{\pi} = \frac{5 }{12} \cdot 180^o =\frac{5 \cdot 180}{12}^o =5(15)^o=75^o \\ \text{ and } 75^o=30^o+45^o\]
could I have done that with the 17pi/12 too or would that be too hard ?
\[\tan(30^o+45^o)=\frac{\tan(30^o)+\tan(45^o)}{1-\tan(30^o)\tan(45^o)}\]
you could have
\[\frac{17 \pi}{12} \cdot \frac{180^o}{\pi}=17(15)^o=255^o \\ \text{ and then instead use } \\ 255^o=225^o+30^o\]
or you could have chosen 120+135 for 255
there or other choices you just want to choose the ones that are on the unit circle
do I change the degrees into the points ? Like from the unit circle
you mean change tan(30) and tan(45) using unit circle? if so yes
yeah , that's what I mean . And sin tan is sin/cos it'd be tan(30) = \[\frac{ \frac{ 1 }{ 2 } }{ \frac{ \sqrt{3} }{ 2 } }\]
thats right
which can be written as 1/sqrt(3) or sqrt(3)/3
SO I'd multiply the denominator and numerator by 2/rad(3) ?
to get that
yep
and you can rationalize the denominator if you prefer which is how I got also or sqrt(3)/3 but you can leave as 1/sqrt(3) for now if you prefer
and tan(45)=?
it'd be 1 right ?
yes tan(45) is 1 \[\tan(30^o+45^o)=\frac{\tan(30^o)+\tan(45^o)}{1-\tan(30^o)\tan(45^o)} \\ \text{ inserting values we just found: } \\ \tan(30^o+45^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}(1)} \\ \tan(75^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}}\] or we could say: \[\tan(\frac{17\pi}{12})=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}}\]
so your teacher probably doesn't want you to leave your answer in compound fraction from
that is tiny fractions inside a big fraction
you can chose to clear the "tiny fractions" by multiply the denominator and numerator of the big fraction by the denominators of the "tiny fractions" so you only need to get rid of the compound action by multiply top and bottom of main fraction by sqrt(3)
\[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)-\sqrt{3}(\frac{1}{\sqrt{3}})} \]
\[\sqrt{3}(1)=\sqrt{3} \\ \text{ and } \\ \sqrt{3} \cdot \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3}}=1\]
couldn't I multiply by 1+sqrt 3 ?
and then use foil
wait you mean 1+1/sqrt(3)?
I rationalized the denominator to just have the sqrt 3
we can rationalize the denominator after what I have... or if you prefer to rationalize first then get rid of the compound fraction action you may do that but you will want to multiply top and bottom by the conjugate of what we have on the bottom which is going to be 1+1/sqrt(3)
What I have right now is \[\frac{ 1+\sqrt{3} }{ 1-\sqrt{3} }\]
that is close to what you should have your bottom should be sqrt(3)-1
why is -1 ?
why is it *
tan(30)=1/sqrt(3) and tan(45)=1 \[\tan(30^o+45^o)=\frac{\tan(30^o)+\tan(45^o)}{1-\tan(30^o)\tan(45^o)} \\ \text{ inserting values we just found: } \\ \tan(30^o+45^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}(1)} \\ \tan(75^o)=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \] \[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)-\sqrt{3}(\frac{1}{\sqrt{3}})} \\ \tan(\frac{17\pi}{12})=\frac{1+\sqrt{3}}{\sqrt{3}-1}\]
into rationalize the denominator we multiply top and bottom by sqrt(3)+1
oooh okay , I rationalized the 1/sqrt 3 before plugging it all in
ok and you winded up with sqrt(3)/3 when you did that right?
\[\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{1 \cdot \sqrt{3}}{ \sqrt{3} \cdot \sqrt{3}}=\frac{\sqrt{3}}{3}\]
this does not equal just sqrt(3)
lol oops , I just canceled them out
\[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)-\sqrt{3}(\frac{1}{\sqrt{3}})} \\ \tan(\frac{17\pi}{12})=\frac{1+\sqrt{3}}{\sqrt{3}-1} \] so basically one last step you multiply top and bottom by sqrt(3)+1
\[\sqrt{3}+\sqrt{3} = 2\sqrt{3}\] ??
so it would be \[2+\sqrt{3}\]
for the final answer
\[\frac{1+\sqrt{3}}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{(1+\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3})^2-1}\] \[\frac{1+\sqrt{3}+\sqrt{3}+3}{3-1}=\frac{4+2 \sqrt{3}}{2}= \frac{4}{2}+\frac{2}{2} \sqrt{3} \\ =2+1 \sqrt{3}=2+\sqrt{3}\] yes you are right
thank you ! I get this a lot more than the last problems lmao
cool stuff!:)
what about sec(pi/8) ?

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