Trig Question

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The first solution set is redundant because all such solutions are contained in the second. The last is incorrect. pi/6 does not work, for instance. For the other solutions you should get cos(x/2) + cos(5x/2) = 0 so 2 * cosx * cos(3x/2) = 0.

Ok so 2npi/5 is the valid one for the solution only?
And for the second solution 2cos(x)cos(3x/2) = 0
=> cos(x)cos(3x/2) = 0 so cosx = 0 and cos3x/2 = 0?

Reply 21

math42

Original post by Fatts13

Ok so 2npi/5 is the valid one for the solution only?
And for the second solution 2cos(x)cos(3x/2) = 0
=> cos(x)cos(3x/2) = 0 so cosx = 0 and cos3x/2 = 0?

Well 4npi/5 is valid but unnecessary
Indeed. Well, either can be 0 (just clarifying that it's or, not and)

Reply 22

Fatts13

Original post by 1 8 13 20 42

Well 4npi/5 is valid but unnecessary
Indeed. Well, either can be 0 (just clarifying that it's or, not and)

OMD thank you so very very much, you've been a massive help :smile:

Sorry for being a hassle.

I'll rep you once TSR let me XD

Reply 23

math42

Original post by Fatts13

OMD thank you so very very much, you've been a massive help :smile:

Sorry for being a hassle.

I'll rep you once TSR let me XD

No worries. Not really a hassle as I am just sitting around today anyway (uni starts tomorrow).

Reply 24

Fatts13

Original post by 1 8 13 20 42

No worries. Not really a hassle as I am just sitting around today anyway (uni starts tomorrow).

If thats the case, here's another question :biggrin:

Reply 25

math42

Original post by Fatts13

If thats the case, here's another question :biggrin:

first one I got a weird general solution but threw in n = 17 and it worked so I'll assume I did it right. Express everything in terms of sin and cos, remove the denominators, and it should all become clear..

Second one I have an expression in tan(theta), it's only slightly simpler. Play around with it, express everything in terms of sintheta and costheta again.
edit: lol it's way simpler just needed to cancel stuff

Reply 26

Fatts13

Original post by 1 8 13 20 42

So tan3theta would become sin3theta/cos3theta and cot(7theta +pi/7) would become cos(7theta+pi/7)/sin(7theta + pi/7)?
And then from there I did:
cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7) and then opened up sin as well and I'm at loss XD

Reply 27

math42

Original post by Fatts13

No need to expand like that. Just cross-multiply, get everything on one side and you should recognise the resulting structure.

Reply 28

Fatts13

Original post by 1 8 13 20 42

No need to expand like that. Just cross-multiply, get everything on one side and you should recognise the resulting structure.

so sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7)?
then does it become tan(3theta)tan(7theta +pi/7)?

Reply 29

math42

Original post by Fatts13

so sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7)?
then does it become tan(3theta)tan(7theta +pi/7)?

I dunno what is going on there..

You have

\displaystyle \frac{ \sin 3 \theta}{ \cos 3 \theta} = \frac{ \cos (7 \theta + \frac{\pi}{7})}{ \sin (7 \theta + \frac{\pi}{7})}

just cross multiply to get everything off the denominator

Reply 30

Fatts13

Original post by 1 8 13 20 42

I dunno what is going on there..

You have

\displaystyle \frac{ \sin 3 \theta}{ \cos 3 \theta} = \frac{ \cos (7 \theta + \frac{\pi}{7})}{ \sin (7 \theta + \frac{\pi}{7})}

just cross multiply to get everything off the denominator

OMD sorry I meant equals not divide.
So sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
I have no clue what to do next :s-smilie:

Reply 31

math42

Original post by Fatts13

OMD sorry I meant equals not divide.
So sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
I have no clue what to do next :s-smilie:

Cos(A + B) = ?

Reply 32

Fatts13

Original post by 1 8 13 20 42

Cos(A + B) = ?

Oh so now we expand? cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7)

Reply 33

math42

Original post by Fatts13

Oh so now we expand? cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7)

Noo; don't complicate, simplify. Try getting everything on one side of the equation (i.e. something = 0) and it might click..

Reply 34

Fatts13

Original post by 1 8 13 20 42

Noo; don't complicate, simplify. Try getting everything on one side of the equation (i.e. something = 0) and it might click..

My head is gone, it's not clicking anything, been deprived of maths for too long.
I got sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7) =1
I honestly don't know how to use cos(A+B) without expanding.

Reply 35

math42

Original post by Fatts13

You had sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
so cos(3theta)cos(7theta+pi/7) - sin(3theta)sin(7theta+pi/7) = 0
Isn't this structure familiar?

Reply 36

Fatts13

Original post by 1 8 13 20 42

You had sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
so cos(3theta)cos(7theta+pi/7) - sin(3theta)sin(7theta+pi/7) = 0
Isn't this structure familiar?

Nope I did not know we could do that. I presumed because everything was multiplication so you could only divide and multiply.
So from there we then have:
cos(3theta + 7theta + pi/7) = 0
cos(10theta + pi/7) = 0
cos^-1(0) = pi/2
10theta + pi/7 = pi/2 + 2npi
so then the general solutions are: +/- pi/28 + pin/5? Correct?