Question

Find the exact value by using a half-angle identity.

tangent of seven pi divided by eight.

Answer 1

first write the half- angle identities down

Answer 2

and recall tan(x) = sin(x) / cos(x)

Answer 3

finally some help.

Answer 4

finally a question that is not solve y=mx+b or simple arithmatic

Answer 5

Lol

Answer 6

\[\cos^2(u) = \frac{ 1+\cos(2u) }{ 2 }\]
and
\[\sin^2(u) = \frac{ 1-\cos(2x) }{ 2 }\]

Answer 7

Ok, give me a min.

Answer 8

so

Answer 9

\[\tan^2(u) = \frac{ \sin^2(u) }{ \cos^2(u) }\]

Answer 10

I would just divide the two identities.

Answer 11

u = 7pi/8 in this problem

Answer 12

what you have so far?

Answer 13

Find all solutions to the equation.

sin2x + sin x = 0

Answer 14

wait

Answer 15

\[\tan^2(u) = \frac{ 1-\cos(2u) }{ 1+\cos(2u) }\]

Answer 16

Hold on.

Answer 17

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

Answer 18

k

Answer 19

recall cos(2u) = cos(2*7pi/8) = cos(7pi/4) =

\[\frac{ \sqrt{2} }{ 2 }\]

Answer 20

\[\sqrt{\frac{ 1-\frac{ \sqrt{2} }{ 2 } }{ 1+\frac{ \sqrt{2} }{ 2 } }}\]

Answer 21

\[\tan(x) = \frac{\sin(x)}{\cos(x)} \implies \tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)}\]\[sin^2(x) = \frac{1}{2}(1-\cos(2x))\]\[\cos^2(x) = \frac{1}{2}(1+cos(2x))\]\[\large\tan^2(x)= \frac{\frac{1}{2}(1-\cos(2x))}{\frac{1}{2}(1+\cos(2x))}\]

Answer 22

yeah i just did that

Answer 23

That's what I was trying to type up, haha,

Answer 24

it is the negative root of the last thing i typed

Answer 25

ok wait, I get it. But im still confused.

Answer 26

which part?

Answer 27

\[\large -\sqrt{\frac{ 1-\frac{ \sqrt{2} }{ 2 } }{ 1+\frac{ \sqrt{2} }{ 2 } }}\]

Answer 28

ok well... Dude, I really dislike this class.

Answer 29

yes

Answer 30

yeah it is correct, both tan 7pi/4 and the root we found is approx -0.414

Answer 31

it is just screwin around with the 5 or 10 identities you learn

Answer 32

no

Answer 33

give me a min, to solve.

Answer 34

Ok I give up, I'm have to really study this some other time. Sorry