Question

Please help me solve this problem!!!? find the exact value of tan 22.5 using the half angle identity?

Answer 1

45

Answer 2

ok i know that i times it by 2 and then ?

Answer 3

tanθ=2tan(θ/2)1+tan2(θ/2)

Answer 4

is that the hanlf angle formula?

Answer 5

i know i am supposed to get \[\sqrt{2}-3 \]

Answer 6

but idk how :/

Answer 7

well, 22.5 is half of 45, right?

Answer 8

yes

Answer 9

the half angle formula is
\[\tan^2( u) = \frac{1-\cos(2u)}{1+\cos(2u)}\]so if we set \(u = 22.5^\circ\), \(2u= 45^\circ\) and you can compute the value of \(\cos(45^\circ)\) without any trouble, right?

Answer 10

is their a square root in this formula?

Answer 11

now, that will give you \(\tan^2 (22.5^\circ)\) and you just take the square root to get \(\tan(22.5^\circ)\)

Answer 12

by the way, \(\tan(22.5^\circ)\) is not \(\sqrt{2} - 3\) but rather \(\sqrt{2}-1\)

Answer 13

oh ok i wrote that wrong

Answer 14

thank you :D

Answer 15

ok i am still a little confused on getting the answer

Answer 16

what do you have so far?

Answer 17

i have \[\sqrt(1}+\sqrt{2/2}\frac{ 2/2 }{-\sqrt{2/2 ? }\]

Answer 18

do you perhaps mean

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

Answer 19

yes

Answer 20

okay, so rationalize the denominator by multiplying both numerator and denominator by the conjugate of the denominator, which is just the denominator with the sign of the radical flipped

Answer 21

you're setting up a difference of squares...

Answer 22

\[1-\sqrt{2}\]

Answer 23

no...

denominator is \[1+\frac{\sqrt{2}}2\]conjugate is \[1-\frac{\sqrt{2}}2\]

\[(a+b)(a-b) = a^2-b^2\]let \[a =1\]\[b=\frac{\sqrt{2}}{2}\]\[(a-b)(a+b) = (1)^2-(\frac{\sqrt{2}}2)^2 = 1 - \frac{2}{4} = \frac{1}{2}\]

so the denominator of the rationalized fraction is 1/2

Answer 24

numerator is \[(1-\frac{\sqrt{2}}2)^2\]giving us

\[\tan^2(22.5^\circ) = \frac{(1-\frac{\sqrt{2}}{2})^2}{\frac{1}{2}}=\]

Answer 25

and then you take the square root of that

\[\sqrt{\tan^2(22.5^\circ) }= \sqrt{\frac{(1-\frac{\sqrt{2}}{2})^2}{\frac{1}{2}}} =\frac{1-\frac{\sqrt{2}}2}{\sqrt{\frac{1}{2}}} = \sqrt{2}(1-\frac{\sqrt{2}}{2})\]

Answer 26

alright thanks i think i understood it aliitle bit more goodnight