anonymous
  • anonymous
Find the exact value of ....
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
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anonymous
  • anonymous
\[\tan (\frac{ 17\pi }{ 12 })\]
freckles
  • freckles
how many 12's are in 17?
anonymous
  • anonymous
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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More answers

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

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