Question

Half Angle Identity:
tan7/12

Answer 1

@calculusfunctions No worries man, if you have to leave, by all means don't worry about me. I'll find help on this material.

Answer 2

Do you mean\[\tan \frac{ 7\pi }{ 12 }\]

Answer 3

It doesn't show that in my notes but that would make more sense

Answer 4

Half-angle identity:\[\tan \frac{ \theta }{ 2 }=\sqrt{\frac{ 1-\cos \theta }{ 1+\cos \theta }}\]

Answer 5

tan(105) = \[\tan(\alpha/2)\]

Answer 6

so \[\alpha = 210\]

Answer 7

Yes\[\frac{ 7\pi }{ 12 }=105° \]so that θ or α = 210° as you said. Very good!

Answer 8

\[\tan(\alpha/2)=\sqrt{\frac{ 1-\cos(210) }{ 1+\cos(210) }}\]

Answer 9

Excellent! Now simplify.

Answer 10

\[-2-\sqrt{3}\] ??

Answer 11

Sorry kinda guessed at that

Answer 12

Nope, Try again. Remember that cos (210°) = -(√3)/2

Answer 13

\[2+\sqrt{3}\]

Answer 14

No, do you want me to type out the steps so that you can study them later?

Answer 15

Yes, please, I'm not seeing this part

Answer 16

\[-\sqrt{\frac{ 1-\cos (210°) }{ 1+\cos (210°) }} \]\[=-\sqrt{\frac{ 1-(\frac{ -\sqrt{3} }{ 2 }) }{ 1+(\frac{ -\sqrt{3} }{ 2 }) }}\]\[=-\sqrt{\frac{ 1+\frac{ \sqrt{3} }{ 2 } }{ 1-\frac{ \sqrt{3} }{ 2 } }}\]\[=-\sqrt{\frac{ \frac{ 2+\sqrt{3} }{ 2 } }{ \frac{ 2-\sqrt{3} }{ 2 } }}\]\[=-\sqrt{\frac{ 2+\sqrt{3} }{ 2-\sqrt{3} }}\]\[=\sqrt{\frac{ (2+\sqrt{3}) }{ (2-\sqrt{3}) }⋅\frac{ (2+\sqrt{3}) }{ (2+\sqrt{3}) }}\]\[=-\sqrt{\frac{ 4+4\sqrt{3}+3 }{ 4-3 }} \]\[=-\sqrt{\frac{ 7+4\sqrt{3} }{ 1 }}\]\[=-\sqrt{7+4\sqrt{3}} \]

Answer 17

The reason for the negative is because 105° is in the second quadrant where tangent is negative.

Answer 18

I hope that helps. I have to go so catch you later.

Answer 19

Thanks a ton for all the help, it makes a lot more sense now. Later.

Answer 20

Later. Glad to know you understand.