Question

Find exact value using half-angle identity cos7pi/12

Answer 1

\(\bf \cfrac{7}{12}\cdot 2\implies \cfrac{7}{6}\qquad thus
\\ \quad \\
cos\left(\cfrac{7\pi}{12}\right)\implies cos\left(\cfrac{{\color{blue}{ \frac{7\pi}{6}}}}{2}\right)=\pm \sqrt{\cfrac{1+cos\left({\color{blue}{ \frac{7\pi}{6}}}\right)}{2}}\)

Answer 2

Is that the final answer? When input that in my calculator I get a decimal answer. But thank you for helping

Answer 3

check your Unit Circle, what's the \(\bf cos\left(\frac{7\pi}{6}\right)\quad ?\)

Answer 4

-sqrt3/2

Answer 5

thus \(\bf cos\left(\cfrac{{\color{blue}{ \frac{7\pi}{6}}}}{2}\right)=\pm \sqrt{\cfrac{1+cos\left({\color{blue}{ \frac{7\pi}{6}}}\right)}{2}}\\ \quad \\\implies cos\left(\cfrac{{\color{blue}{ \frac{7\pi}{6}}}}{2}\right)=\pm \sqrt{\cfrac{1+\left({\color{blue}{ -\frac{\sqrt{3}}{2}}}\right)}{2}}\)

and then simplify the radicand

Answer 6

Yay! Ok that's what I have so far! But when I punch that in the calculator I still get a stupid decimal :((

Answer 7

yes, the calculator will give you the decimal value
so that's correct, if you want the decimal value

if you just want the simplified version, you'd need to simplify the fractions in the root
to make them just 1 fraction

Answer 8

So do I multiply the bottom 2 to the top

Answer 9

perform the SUM in the numerator, and keep in mind that \(\bf \cfrac{\quad \frac{a}{b}\quad }{\frac{c}{d}}\implies \cfrac{a}{b}\cdot \cfrac{d}{c}\)

Answer 10

I'm sorry I'm a total airhead when it comes to math

Answer 11

So multiply by the reciprocal :)

Answer 12

yes, because \(\bf cos\left(\cfrac{ \frac{7\pi}{6}}{2}\right)=\pm \sqrt{\cfrac{1+\left({\color{blue}{ -\frac{\sqrt{3}}{2}}}\right)}{2}}
\implies
cos\left(\cfrac{ \frac{7\pi}{6}}{2}\right)=\pm \sqrt{\cfrac{1 -\frac{\sqrt{3}}{2}}{\frac{2}{1}}}\)

Answer 13

so... what would you get for \(\bf 1 -\cfrac{\sqrt{3}}{2}\) ?

Answer 14

Ok so far after that I have sqrt1+cos-sqrt3 and that's inside a big sqrt

Answer 15

hmmm recall your fractions additions \(\bf cos\left(\cfrac{ \frac{7\pi}{6}}{2}\right)=\pm \sqrt{\cfrac{1+\left({\color{blue}{ -\frac{\sqrt{3}}{2}}}\right)}{2}}
\implies
cos\left(\cfrac{ \frac{7\pi}{6}}{2}\right)=\pm \sqrt{\cfrac{1 -\frac{\sqrt{3}}{2}}{\frac{2}{1}}}
\\ \quad \\
cos\left(\cfrac{ \frac{7\pi}{6}}{2}\right)=\pm \sqrt{\cfrac{\frac{2-\sqrt{3}}{2}}{\frac{2}{1}}}\)

Answer 16

so you'd have two fractions, so recall \(\bf \cfrac{\quad \frac{a}{b}\quad }{\frac{c}{d}}\implies \cfrac{a}{b}\cdot \cfrac{d}{c}\)

Answer 17

Doesn't the bottom cancel out?

Answer 18

hmmmm flip the denominator, let's see if it does cancel out

Answer 19

Not if I flip it because then it's 1/2
:/

Answer 20

\(\bf \cfrac{\frac{2-\sqrt{3}}{2}}{\frac{2}{1}}\implies \cfrac{2-\sqrt{3}}{2}\cdot \cfrac{1}{2}\)

Answer 21

Omg I forgot the two before the -sqrt3

Answer 22

\(\bf cos\left(\cfrac{ \frac{7\pi}{6}}{2}\right)=\pm \sqrt{\cfrac{1+\left({\color{blue}{ -\frac{\sqrt{3}}{2}}}\right)}{2}}
\implies
cos\left(\cfrac{ \frac{7\pi}{6}}{2}\right)=\pm \sqrt{\cfrac{1 -\frac{\sqrt{3}}{2}}{\frac{2}{1}}}
\\ \quad \\
cos\left(\cfrac{ \frac{7\pi}{6}}{2}\right)=\pm \sqrt{\cfrac{\frac{2-\sqrt{3}}{2}}{\frac{2}{1}}}\implies cos\left(\cfrac{ \frac{7\pi}{6}}{2}\right)=\pm\sqrt{\cfrac{2-\sqrt{3}}{4}}\)

Answer 23

That looks so much more clear. Thank you so so so much !!!! :)

Answer 24

yw