Question

sec^2 (pi/5) - tan^2 (pi/5)

Answer 1

noted... thanks for sharing that

Answer 2

oh ok do you know your Pythagorean identities?

Answer 3

sin ^2 + cos ^2 = 1
i know theres 3 more but i forgot :/

Answer 4

you can get the others by dividing that equation by cos^2
or dividing it by sin^2

let's just divide both sides by cos^2 for now

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

Answer 5

so my problem = 1?

Answer 6

yep

Answer 7

wow i have 1 more im also confused on

Answer 8

k what is it

Answer 9

\[(\cos 2\pi/9)(\sec 2\pi/9)\]

Answer 10

cosine and secant are reciprocals of one another

Answer 11

\[\sec(x)=\frac{1}{\cos(x)}\]

Answer 12

oh yeah! so it = 1 lol

Answer 13

yes

Answer 14

im so done....plus i never got taught those identities yet

Answer 15

what is this? like itt?

Answer 16

like college?

Answer 17

im in high school precalc honors

Answer 18

ah

Answer 19

well i think in this is my opinion teachers shouldn't have to teach everything

Answer 20

just the basics and then students can use those basics to make their own discoveries

Answer 21

I also couldn't find the examples in the book

Answer 22

to help solve for those 2

Answer 23

well now you have them
examples that is

Answer 24

thank you