Question

identities question to find values of x below

Answer 1

\[\cos 4x=-\frac{ \sqrt{2} }{ 2 }\]

Answer 2

I would probably start with doing a sum of cosines and doing cos(2x + 2x) and see where that starts you off.

Answer 3

@Psymon that would give me a decimal for my answer. I believe the question requires me to write in all radians without the use of a calculator. how would you take the cos(2) without a calculator?

Answer 4

\(\bf cos (4x)=-\cfrac{ \sqrt{2} }{ 2 }\\
cos^{-1}(cos(4x))=cos^{-1}\left(-\cfrac{ \sqrt{2} }{ 2 }\right)\\
4x = cos^{-1}\left(-\cfrac{ \sqrt{2} }{ 2 }\right)\)

Answer 5

\(\bf 4x = cos^{-1}\left(-\cfrac{ \sqrt{2} }{ 2 }\right) \implies
4x = \text{some angle} \implies x = \cfrac{\text{some angle}}{4}\)

Answer 6

Yeah, he knows what he's doing more than I am, ignore me x_x

Answer 7

@Psymon that's ok @jdoe0001 how can i solve the inverse of cosine without a calculator? The question requires me to use radians only

Answer 8

I see identities and thats where my mind goes. Depending on your calculator, putting in cos^(-1)sqrt2/2 will give you the radian answer. Otherwise it's unit circle knowledge. cos^(-1)sqrt2/2 is a clean match with 2 radian measures : )

Answer 9

\(\bf -\cfrac{ \sqrt{2} }{ 2 }\) is a pretty well known angle, it should be in your Unit Circle

Answer 10

Sounds good to me :P

Answer 11

sorry, it got 3pi/16and 5pi/16

Answer 12

well, you have the correct "reference angle", but keep in mind that in this case cosine is negative, that means 2nd and 3rd quadrants

Answer 13

O.o For cosine?. That would mean that you started with 3pi/4 and 5pi/4 = 4x, but those are negative in those quadrants. The first answer you had was correct.

Answer 14

3pi/16and 5pi/16 :)

Answer 15

Oh, I thought it was positive sqrt2/2, didnt notice the negative x_x

Answer 16

yea i didn't consider the negative for my first answer. I got @jdoe0001 's answer after i looked at the negative quadrants

Answer 17

Yeah, I just simply see the negative. I swear im horrible at reading math when its in typing O.o

Answer 18

*didnt see

Answer 19

hheehe, sneaky dash eludes at times too, since it's so small "-"

Answer 20

Exactly. But nah, otherwise you got the right answer, haha. Although Im surprised they wouldnt have you memorize the unit circle, morning.

Answer 21

@Psymon and @jdoe0001 thank you so much! may i ask one more question?? it's been bugging me for at least 7 hours!

Answer 22

If my brain decides to work, haha.

Answer 23

@Psymon i'm sure it will work where mine fails :)

Answer 24

Go for it then. Oh, and how do you kinda quote people like that, never knew how it does that.

Answer 25

here it is: how would you solve \[\tan^{-1} 6 without a calculator?/\]

Answer 26

inverse tan of 6, right?

Answer 27

yup

Answer 28

\(\bf tan^{-1}(6) \ \ ?\)

Answer 29

yes :)

Answer 30

i wanted to use a sum and difference formula but i got crazy radians

Answer 31

Yeah, not even sure how you would do it without something crazy like a sum formula. 6 is no exact number for anything inverse. I just cant imagine doing that without some ridiculous identity process. I could be wrong, though

Answer 32

:( exactly. the original problem is actually \[\cos (\frac{ 1 }{ 2 }\tan^{-1} 6\]

Answer 33

cos[ (1/2)*tan^(-1)6 ] ? Sorry, that math writing makes no sense to me xD

Answer 34

that's why i'm here lol I know that i'm supposed to solve the inverse of tan first. and then half that answer.

Answer 35

Not quite, lol. But did I get the problem correct?

Answer 36

The original problem? yes you did :) thank you. I'f you'd like to opt out of this new problem that's ok :) I can re-post it just in case someone else can explain it . Thank you so much to you @Psymon and @jdoe0001

Answer 37

Well, I have a kinda sorta idea.......xD

Answer 38

Wanna hear me out and see if it makes any sense?

Answer 39

No idea if this works, but @morningperson1991 not sure if you wanted to hear me out on this one.

Answer 40

that's fine :) I'll post it in a few minutes . Thank you for all that you do!