Question

If you help me, i'll give you a medal :D
Find the exact value by using a half-angle identity. tan(7pi/8)

Answer 1

$$\bf
\cfrac{7\pi}{8} \times 2 \implies \cfrac{7\pi}{4}\\
\cfrac{\frac{7\pi}{4}}{2} \implies \cfrac{7\pi}{8}\\
tan\left(\cfrac{\frac{7\pi}{4}}{2}\right) =
\cfrac{sin\left(\frac{7\pi}{4}\right)}{1+cos\left(\frac{7\pi}{4}\right)}
$$
so, what are the values for sine and cosine for \(\bf \cfrac{7\pi}{4}\ \ ?\)

Answer 2

sin is sqrt 2 and cos is - sqrt 2

Answer 3

\(\bf \cfrac{7\pi}{4}\) is in the 4th quadrant
in the 4th quadrant sine is negative and cosine is positive, otherwise the values are correct

Answer 4

or you can use this Unit Circle instead -> http://i.stack.imgur.com/r8uHr.gif

Answer 5

ooooooooooh I forgot that part :p

Answer 6

\(\bf tan\left(\cfrac{\frac{7\pi}{4}}{2}\right) =
\cfrac{sin\left(\frac{7\pi}{4}\right)}{1+cos\left(\frac{7\pi}{4}\right)}\\
tan\left(\cfrac{\frac{7\pi}{4}}{2}\right) =\large \cfrac{-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}\)

Answer 7

what would that give you?

Answer 8

- sqrt 2/ 1 + 2sqrt2?

Answer 9

yes, as you'd noticed in the Unit Circle, those are the values for cosine and sine at \(\bf \cfrac{7\pi}{4}\)

Answer 10

Ok, so that would be the answer?

Answer 11

well, what does \(\bf \large \cfrac{-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}\) give you?

Answer 12

I'm not sure. Do you have to change the 1?

Answer 13

keep in mind that, in the denominator => \(\bf \large 1+\frac{\sqrt{2}}{2} \implies \frac{1}{1}+\frac{\sqrt{2}}{2}\) , so is just a plain fractions addition

Answer 14

so it's \[-\sqrt{2}/ ((2+\sqrt{2})/ 2)\]

Answer 15

well, the numerator should be negative, as in \(\bf -\cfrac{\sqrt{2}}{2}\)
the denominator is fine though

Answer 16

\(\bf \large \bf \cfrac{-\frac{\sqrt{2}}{2}}
{\frac{2+\sqrt{2}}{2}} \implies -\frac{\sqrt{2}}{2} \times \frac{2}{2+\sqrt{2}}
\)

Answer 17

\[-(2\sqrt{2})/ 4 +2\sqrt{2}\]

Answer 18

well \(\bf \large -\frac{\sqrt{2}}{\cancel{2}} \times \frac{\cancel{2}}{2+\sqrt{2}} \implies \cfrac{-\sqrt{2}}{2+\sqrt{2}}\)

Answer 19

Thank you! :D

Answer 20

one sec

Answer 21

$$\bf \text{now simplifying the denominator by multiplying for the conjugate}\\
\cfrac{-\sqrt{2}}{2+\sqrt{2}} \times \cfrac{2-\sqrt{2}}{2-\sqrt{2}}\\
\text{keep in mind that}\\
(a-b)(a+b) = (a^2-b^2)\\
\cfrac{-\sqrt{2} \times (2-\sqrt{2})}{(2)^2-(\sqrt{2})^2}
$$

Answer 22

$$\bf \frac{2-2\sqrt{2}}{4-2} \implies \frac{\cancel{2}-\cancel{2}\sqrt{2}}{\cancel{2}}
$$

Answer 23

so, that leaves you with a simplified version of \(\bf 1-\sqrt{2}\)

Answer 24

Oooooooooooooooooooh! Thank you!!!

Answer 25

yw