Question

find the exact value of tan22.5 using the half angle formula

Answer 1

so... what's 22.5 * 2?

Answer 2

45

Answer 3

\(\bf tan(22.5^o)=tan\left(\frac{22.5^o\cdot 2}{2}\right)=tan\left(\frac{45^o}{2}\right)\)

so there, use that in the half-angle identity for tangent :)

Answer 4

okay at that point what is the next step to be taken

Answer 5

using the trig half-angle identity :)

Answer 6

there are about 3 for tangent, so pick either

Answer 7

is it \[\sqrt{1-\cos45}/\sqrt{1+\cos45}\]

Answer 8

yeap

Answer 9

\(\bf tan(22.5^o)=tan\left(\frac{22.5^o\cdot 2}{2}\right)=tan\left(\frac{45^o}{2}\right)=\pm\sqrt{\cfrac{1-cos(45^o)}{1+cos(45^o)}}
\)

Answer 10

and that's a common known angle, and the cosine would be in your Unit Circle

Answer 11

so cos 45 is \[\sqrt{2}/2\] how do i finish it im not great with the algebra

Answer 12

ok... one sec

Answer 13

k

Answer 14

\(\bf tan\left(\frac{45^o}{2}\right)=\pm\sqrt{\cfrac{1-cos(45^o)}{1+cos(45^o)}}\\ \quad \\
\implies
\pm\sqrt{\cfrac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}}\qquad \textit{let us do the inside first}\implies \cfrac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}\\ \quad \\
\cfrac{\frac{2-\sqrt{2}}{2}}{\frac{2+\sqrt{2}}{2}}\implies
\frac{2-\sqrt{2}}{2}\cdot \frac{2}{2+\sqrt{2}}\implies \cfrac{2-\sqrt{2}}{2+\sqrt{2}}\\ \quad \\
\textit{multiplying by the conjugate of the denominator}\\ \quad \\
\cfrac{2-\sqrt{2}}{2+\sqrt{2}}\cdot \cfrac{2-\sqrt{2}}{2-\sqrt{2}}\implies \cfrac{(2-\sqrt{2})^2}{2^2-\sqrt{2^2}}\\ \quad \\
\cfrac{4-4\sqrt{2}+\sqrt{2^2}}{4-2}\implies \cfrac{4-4\sqrt{2}+2}{2}\implies

\cfrac{\cancel{6}-\cancel{4}\sqrt{2}}{\cancel{2}}\\ \quad \\

3-2\sqrt{2}\Leftarrow\textit{that's the inside, thus}\\ \quad \\

tan\left(\frac{45^o}{2}\right)\sqrt{3-2\sqrt{2}}\)

Answer 15

hmm well \(\bf tan\left(\frac{45^o}{2}\right)=\pm\sqrt{3-2\sqrt{2}}\) that is

Answer 16

thank you