Fanduekisses
  • Fanduekisses
Do I use the half angle formula to solve tan(5pi/12) ????
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Fanduekisses
  • Fanduekisses
\[\tan \frac{ 5\pi }{ 12 }=\frac{ \sqrt{1/2(1-\cos(2*\frac{ 5\pi }{ 12 })} }{ \sqrt{1/2(1+\cos(2*\frac{ 5\pi }{ 12 })} }\]
Fanduekisses
  • Fanduekisses
like that? Is at least my approach good?
freckles
  • freckles
that is one way to go about it or you could use that 5pi/12 is equal to pi/6+pi/4 and use sum identity which ever one you prefer here

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Fanduekisses
  • Fanduekisses
Great, so it's a matter of preference or efficiency ;) thanks! <3
Fanduekisses
  • Fanduekisses
I get a nicer number using the sum identity lol. I get 1.
Fanduekisses
  • Fanduekisses
With the half angle identity I get 3.732... lol
freckles
  • freckles
well some mistake must have occurred let's see ...
freckles
  • freckles
\[\sin(\frac{1}{2} \cdot \frac{ 5 \pi}{6})=\sqrt{\frac{1-\cos(\frac{5 \pi}{6})}{2}} \\ \cos(\frac{1}{2} \cdot \frac{5 \pi}{6})=\sqrt{\frac{1+\cos(\frac{5\pi}{6})}{2}} \\ \tan(\frac{5\pi}{12}) =\frac{\sqrt{\frac{1-\cos(\frac{5\pi}{6})}{2}}}{\sqrt{\frac{1+\cos(\frac{5\pi}{6}}{2}}}=\frac{\sqrt{1-\cos(\frac{5\pi}{6})}}{\sqrt{1+\cos(\frac{5\pi}{6})}}\] what did you get for cos(5pi/6)?
Fanduekisses
  • Fanduekisses
-sqrt 3pi/2
freckles
  • freckles
my notifications are working poorly
freckles
  • freckles
don't you mean -sqrt(3)/2 ?
Fanduekisses
  • Fanduekisses
yes lol
freckles
  • freckles
\[\tan(\frac{5\pi}{12})=\frac{\sqrt{1+\frac{\sqrt{3}}{2}}}{\sqrt{1-\frac{\sqrt{3}}{2}}}=\frac{\sqrt{\frac{2+\sqrt{3}}{2}}}{\sqrt{\frac{2-\sqrt{3}}{2}}} =\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}= \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}} \\ \tan(\frac{5 \pi}{12})=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}} \cdot \sqrt{\frac{2+\sqrt{3}}{2+\sqrt{3}}} =\sqrt{\frac{(2+\sqrt{3})^2}{4-3}}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3}\] now by addition method... \[\tan(\frac{\pi}{4}+\frac{\pi}{6})=\frac{\tan(\frac{\pi}{4})+\tan(\frac{\pi}{6})}{1-\tan(\frac{\pi}{4}) \tan(\frac{\pi}{6})}\] is this what you computed by addition method?
freckles
  • freckles
my answer got cut off
freckles
  • freckles
\[\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3}\]
freckles
  • freckles
\[\tan(\frac{\pi}{4})=1 \text{ while } \tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}\]
freckles
  • freckles
\[\tan(\frac{\pi}{4}+\frac{\pi}{6})=\frac{1+\frac{1}{\sqrt{3}}}{1-1 \cdot \frac{1}{\sqrt{3}}} \\ =\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}= \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} =\frac{\sqrt{3}+1}{\sqrt{3}-1}\] try rationalizing the denominator
Fanduekisses
  • Fanduekisses
I used sin and cos sum identity lol
Fanduekisses
  • Fanduekisses
I did \[\frac{ \sin(\frac{ \pi }{ 6 } +\frac{ \pi }{ 4 })}{ \cos (\frac{ \pi }{ 6 }+\frac{ \pi }{ 4 }) }\]
freckles
  • freckles
so you will need to evaluate the following: cos(pi/6) sin(pi/6) cos(pi/4) sin(pi/4)
Fanduekisses
  • Fanduekisses
yes
Fanduekisses
  • Fanduekisses
I did it and I got 1
freckles
  • freckles
you can get these off the unit circle don't put pi in any of the answers
Fanduekisses
  • Fanduekisses
yes cos pi/6 is sqrt(3)/2
freckles
  • freckles
\[\sin(\frac{\pi}{6}+\frac{\pi}{4})=\sin(\frac{\pi}{6}) \cos(\frac{\pi}{4})+\sin(\frac{\pi}{4}) \cos(\frac{\pi}{6}) \\ \sin(\frac{\pi}{6}+\frac{\pi}{4})=\frac{1}{2} \cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} \\ \sin(\frac{\pi}{6}+\frac{\pi}{4})= \frac{\sqrt{2}+\sqrt{6}}{4} \\ \cos(\frac{\pi}{6}+\frac{\pi}{4})=\cos(\frac{\pi}{6} ) \cos(\frac{\pi}{4})-\sin(\frac{\pi}{6})\sin(\frac{\pi}{4}) \\ \cos(\frac{\pi}{6}+\frac{\pi}{4})=\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}-\frac{1}{2} \cdot \frac{\sqrt{2}}{2} \\ \cos(\frac{\pi}{6}+\frac{\pi}{4})= \frac{\sqrt{6}-\sqrt{2}}{4} \\ \tan(\frac{5\pi}{12})=\frac{\sqrt{2}+\sqrt{6}}{\sqrt{6}-\sqrt{2}} \neq 1 \]
freckles
  • freckles
\[\tan(\frac{5 \pi}{12}) =\frac{\sqrt{2} +\sqrt{2} \sqrt{3}}{\sqrt{2} \sqrt{3}-\sqrt{2}} \\ \tan(\frac{5\pi}{12})=\frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{1+\sqrt{3}}{\sqrt{3}-1} \\ \tan(\frac{5\pi}{12}) =\frac{1+\sqrt{3}}{\sqrt{3}-1}\] rationalize the denominator
Fanduekisses
  • Fanduekisses
sqrt(3)+2
Fanduekisses
  • Fanduekisses
I see what my mistake was. I put a plus instead of a minus in the cos sun identity lol
Fanduekisses
  • Fanduekisses
Thanks <3

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