Question

Trig Problems Please Help?!

Answer 1

what r the problems

Answer 2

give me a sec

Answer 3

1. Find the exact value of tan(17π / 12).
2. Prove the identity (sin(2x) / sin x) – (cos(2x) / cos x) = sec x.
3. Solve sin(x + (π / 4)) – sin(x – (π / 4) = 1.

Answer 4

\[\tan \frac{ 17 \pi }{ 12 } \] is same as getting \[\tan \frac{ 5 }{ 12 }\] through reference angle theorem, right?

Answer 5

and you know that \[\tan \frac{ 5 }{ 12 } = \tan (\frac{ 1 }{ 4 }+\frac{ 1 }{ 6 })\] so apply identies

Answer 6

do you follow @B.Sanders

Answer 7

Not really I don't understand this stuff at all.

Answer 8

if you don't understand this stuff at all, i would suggest that these different and complicated questions is not the place to begin

Answer 9

i can walk you through one of them if you like, but my guess is it would not make sense or help in any way

Answer 10

We could try

Answer 11

pick one if you like
by 'this stuff" do you mean these problems or trig in general? these problems have to do with addition angle formulas (and subtraction)

Answer 12

I don't understand problems like these

Answer 13

ok so here is a nice cheat sheet with the formulas we need
then we can use them

Answer 14

now there are a lot of formulas on there, but we are only using the "addition and subtraction" formulas for these problems
do you see them? one says something like
\[\sin(\alpha +\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\]

Answer 15

Yea

Answer 16

ok so lets tackle the last problem first, it is easiest to explain

Answer 17

actually there are a couple ways to do it but lets just use the formula \[\sin(\alpha +\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\] for the first term in problem 3, which is
\[\sin(x+\frac{\pi}{4})\]
i hope it is clear that the way to use the formula is to replace \(\alpha\) by \(x\) and \(\beta\) by \(\frac{\pi}{4}\)
you with me so far?

Answer 18

yep

Answer 19

ok now that is easy for me to do, i can just copy and paste

Answer 20

\[\sin(\alpha +\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\]
\[\sin(x+\frac{\pi}{4})=\sin(x)\cos(\frac{\pi}{4})+\cos(x)\sin(\frac{\pi}{4})\]

Answer 21

clear so far? we can actually evaluate two of these, but there may be no need to do it at this step

Answer 22

Yes

Answer 23

ok now we are going to compute \[\sin(x-\frac{\pi}{4})=\sin(x)\cos(\frac{\pi}{4})-\cos(x)\sin(\frac{\pi}{4})\]

Answer 24

then we are going to subtract the second one from the first
don't forget the distributive law

Answer 25

you get, if my algebra is correct,
\[2\cos(x)\sin(\frac{\pi}{4})\]
look good?

Answer 26

How did you get 2cos(x)sin(pi/4) if cos is paired with x and pi/4 and so is sin?

Answer 27

this is what i did
easier to write without all the sines and cosines
\[A+B-(A-B)=2B\]

Answer 28

let me know if that is clear yet
otherwise i have to write it out with all the sines and cosines, which i guess i can do by copying and pasting

Answer 29

\[\sin(x)\cos(\frac{\pi}{4})+\cos(x)\sin(\frac{\pi}{4})-[\sin(x)\cos(\frac{\pi}{4})-\cos(x)\sin(\frac{\pi}{4})]\]
\[=2\cos(x)\sin(\frac{\pi}{4})\]

Answer 30

the terms \(\sin(x)\cos(\frac{\pi}{4})\) cancel, i.e. add to zero

Answer 31

Okay makes sense

Answer 32

ok that is all the use of the formula
now our job is to solve
\[2\cos(x)\sin(\frac{\pi}{4})=1\] so now it is time to evaluate \(\sin(\frac{\pi}{4})\)
do you know what that is?

Answer 33

One min

Answer 34

0.014?

Answer 35

it is on the cheat sheet i sent you, but if you are going to do a bunch of trig you should memorize this one

Answer 36

I used my calculator the way my teacher showed us. I could be wrong tho.

Answer 37

oh no my dear, lets back up and go slow
first of all forget about decimals
take a look at the unit circle on the last page of the cheat sheet

Answer 38

I am

Answer 39

this is not a calculator exercise in any way
you need the "exact" value
look at the unit circle on the last page of the cheat sheet i sent
find \(\frac{\pi}{4}\) on the circle, then look at the corresponding coordinates
let me know when you see it

Answer 40

I see it the values are \[\frac{ \sqrt{2} }{ 2 },\frac{ \sqrt{2} }{ 2 }\]

Answer 41

ok good!
the first coordinate is \(\cos(\frac{\pi}{4})\)and the second coordinate is \(\sin(\frac{\pi}{4})\) but since they are both the same, you can't really mess this up
now we have
\[\sin(\frac{\pi}{4})=\frac{\sqrt2}{2}\]

Answer 42

now the equation \[2\cos(x)\sin(\frac{\pi}{4})=1\] becomes
\[2\cos(x)\frac{\sqrt2}{2}=1\] or more simply put
\[\sqrt2\cos(x)=1\]

Answer 43

almost done
you got that part?

Answer 44

Sorry got kicked off.