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anonymous
 one year ago
Find the exact value of ....
anonymous
 one year ago
Find the exact value of ....

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\tan (\frac{ 17\pi }{ 12 })\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2how many 12's are in 17?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know I have to use the equation \[\tan(a+b)=(\frac{ \tan(a)\tan(b) }{ 1+\tan(a)\tan(b) })\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2right 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{5\pi}{12} \cdot \frac{180^o}{\pi} = \frac{5 }{12} \cdot 180^o =\frac{5 \cdot 180}{12}^o =5(15)^o=75^o \\ \text{ and } 75^o=30^o+45^o\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0could I have done that with the 17pi/12 too or would that be too hard ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\tan(30^o+45^o)=\frac{\tan(30^o)+\tan(45^o)}{1\tan(30^o)\tan(45^o)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{17 \pi}{12} \cdot \frac{180^o}{\pi}=17(15)^o=255^o \\ \text{ and then instead use } \\ 255^o=225^o+30^o\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2or you could have chosen 120+135 for 255

freckles
 one year ago
Best ResponseYou've already chosen the best response.2there or other choices you just want to choose the ones that are on the unit circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do I change the degrees into the points ? Like from the unit circle

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you mean change tan(30) and tan(45) using unit circle? if so yes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah , 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
 one year ago
Best ResponseYou've already chosen the best response.2which can be written as 1/sqrt(3) or sqrt(3)/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0SO I'd multiply the denominator and numerator by 2/rad(3) ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and 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
 one year ago
Best ResponseYou've already chosen the best response.2yes 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
 one year ago
Best ResponseYou've already chosen the best response.2so your teacher probably doesn't want you to leave your answer in compound fraction from

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that is tiny fractions inside a big fraction

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)\sqrt{3}(\frac{1}{\sqrt{3}})} \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{3}(1)=\sqrt{3} \\ \text{ and } \\ \sqrt{3} \cdot \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3}}=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0couldn't I multiply by 1+sqrt 3 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2wait you mean 1+1/sqrt(3)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I rationalized the denominator to just have the sqrt 3

freckles
 one year ago
Best ResponseYou've already chosen the best response.2we 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
 one year ago
Best ResponseYou've already chosen the best response.0What I have right now is \[\frac{ 1+\sqrt{3} }{ 1\sqrt{3} }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that is close to what you should have your bottom should be sqrt(3)1

freckles
 one year ago
Best ResponseYou've already chosen the best response.2tan(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
 one year ago
Best ResponseYou've already chosen the best response.2into rationalize the denominator we multiply top and bottom by sqrt(3)+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oooh okay , I rationalized the 1/sqrt 3 before plugging it all in

freckles
 one year ago
Best ResponseYou've already chosen the best response.2ok and you winded up with sqrt(3)/3 when you did that right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{1 \cdot \sqrt{3}}{ \sqrt{3} \cdot \sqrt{3}}=\frac{\sqrt{3}}{3}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2this does not equal just sqrt(3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol oops , I just canceled them out

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\tan(\frac{17\pi}{12})=\frac{\sqrt{3}}{\sqrt{3}} \cdot \frac{\frac{1}{\sqrt{3}}+1}{1\frac{1}{\sqrt{3}}} \\ \tan(\frac{17\pi}{12})=\frac{\sqrt{3} \cdot (\frac{1}{\sqrt{3}})+\sqrt{3}(1)}{\sqrt{3}(1)\sqrt{3}(\frac{1}{\sqrt{3}})} \\ \tan(\frac{17\pi}{12})=\frac{1+\sqrt{3}}{\sqrt{3}1} \] so basically one last step you multiply top and bottom by sqrt(3)+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{3}+\sqrt{3} = 2\sqrt{3}\] ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it would be \[2+\sqrt{3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the final answer

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{1+\sqrt{3}}{\sqrt{3}1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{(1+\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3})^21}\] \[\frac{1+\sqrt{3}+\sqrt{3}+3}{31}=\frac{4+2 \sqrt{3}}{2}= \frac{4}{2}+\frac{2}{2} \sqrt{3} \\ =2+1 \sqrt{3}=2+\sqrt{3}\] yes you are right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you ! I get this a lot more than the last problems lmao

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about sec(pi/8) ?
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