Question

Tan7pi/8 using half life formulas, find the exact value. Can someone help me solve this

Answer 1

Hey c:

Answer 2

Tangent Half Angle? Hmm, there are a bunch of different forms of that one, ya?
How bout we use this one?\[\Large\rm \tan\left(\frac{x}{2}\right)=\frac{\sin x}{1+\cos x}\]
So we setup our half angle first:
\[\Large\rm \tan\left(\frac{7\pi}{8}\right)=\tan\left(\frac{7\pi/4}{2}\right)\]And then plug in the stuff,\[\Large\rm =\frac{\sin\left(\frac{7\pi}{4}\right)}{1+\cos\left(\frac{7\pi}{4}\right)}\]

Answer 3

do you know how to compute
sin(7 pi/4)
and
cos(7 pi/4) ?

Answer 4

hint:
\[\Large \begin{gathered}
\sin \left( {\frac{{7\pi }}{4}} \right) = \sin \left( {2\pi - \frac{\pi }{4}} \right) = \hfill \\
= \sin \left( {2\pi } \right)\cos \left( {\frac{\pi }{4}} \right) - \cos \left( {2\pi } \right)\sin \left( {\frac{\pi }{4}} \right) = \hfill \\
= 0 \times \frac{1}{{\sqrt 2 }} - 1 \times \frac{1}{{\sqrt 2 }} = ...? \hfill \\
\end{gathered} \]

Answer 5

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

Answer 6

This is the formula we used

Answer 7

@zepdrix and @Michele_Laino

Answer 8

so, we can write:
\[\Large \tan \left( {\frac{{7\pi }}{8}} \right) = \sqrt {\frac{{1 - \cos \left( {\frac{{7\pi }}{4}} \right)}}{{1 + \cos \left( {\frac{{7\pi }}{4}} \right)}}} \]

Answer 9

Where does the 4 come from?

Answer 10

since if
alpha/2= 7 pi/8,
then
alpha = 7 pi/4

Answer 11

please keep in mind that, the formula that you used, is:
\[\Large \tan \left( {\frac{\alpha }{2}} \right) = \sqrt {\frac{{1 - \cos \alpha }}{{1 + \cos \alpha }}} \]

Answer 12

So we only divide the bottom by 2?

Answer 13

To get 7pi/4

Answer 14

we have to multiply (7 pi/8) by 2, in order to get (7 pi/4)

Answer 15

Can you show me the work you did because I'm still confused

Answer 16

we have:
\[\Large \begin{gathered}
\cos \left( {\frac{{7\pi }}{4}} \right) = \cos \left( {2\pi - \frac{\pi }{4}} \right) = \hfill \\
= \cos \left( {2\pi } \right)\cos \left( {\frac{\pi }{4}} \right) + \sin \left( {2\pi } \right)\sin \left( {\frac{\pi }{4}} \right) = \hfill \\
= 1 \times \frac{1}{{\sqrt 2 }} + 0 \times \frac{1}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \hfill \\
\end{gathered} \]

Answer 17

so, after substitution, we can write:
\[\Large \tan \left( {\frac{{7\pi }}{8}} \right) = \sqrt {\frac{{1 - \frac{1}{{\sqrt 2 }}}}{{1 + \frac{1}{{\sqrt 2 }}}}} \]

Answer 18

Got it, I'm going to use sqrt 2/2 instead of 1/sqrt2